NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression

Arithmetic Progression Introduction

In daily life, there are so many examples which follow a certain pattern.

  1. In a fixed deposit scheme, the amount becomes. 1 item of itself after every year. The maturity amount of ₹10,000 after 1, 2, 3 and 4 years will be ₹11,000, ₹12,100, ₹13,310,14.641
  2. When a person is offered a job with a monthly salary of ₹18,000 and with an annual increment of ₹400, His salary for the succeeding years will be ₹18,000, ₹18,400, ₹ 18,800 per mon.
    • In the first example, the succeeding terms are obtained by multiplying with 1.1. In the second example, the succeeding terms are obtained by adding 400.
    • Now, we will discuss all those patterns in which the succeeding terms are obtained by adding a fixed number to the preceding terms.

Sequence

A number of things that come one after another form a sequence. It may be possible that we do not have a formula to find the nth term of the sequence, but still, we know about the next term.

Sequence For example: 2, 3, 5, 7, 11,13, 17, 19, …is a sequence of prime numbers.

We all know about the next number of the sequence but we do not have any formula to calculate the next number.

  • The numbers present in the sequence are called the terms of the sequence.
  • The nth term of the sequence can be represented by Tn or an. It is called the general term.
  • A sequence which has finite terms is called a finite sequence.

Progression

Those sequences whose terms follow certain patterns are called progressions. In progression, each term (except the first) is obtained from the previous one by some rule.

The difference between a progression and a sequence is that a progression has a specific formula to calculate its nth term, whereas a sequence can be based on a logical rule like a group of prime numbers, which does not have any formula associated with it.

The numbers 2, 4, 6, 8, 10, 12 … form a progression as its nth term Tn =2n while 2,3,5,7,11,13,17,19,23,29,31,… is a sequence of numbers, as there is no formula to find the next number. It is a group of prime numbers.

Series

If T1, T2, T3,…, Tn form a progression, then T1, T2, T3,…, Tn is called its corresponding series. In other words, if all the terms of a progression are connected by ‘+’ started

Series For example:1-2 + 3- 4+5-6 ……… is a series. The first term is 1, the second term is -2, the third term is 3, the fourth is -4 and so on. Its nth is n(-1) n+1

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression

Series Solved Examples

Question 1. The nth of a sequence is an = 2n+1. Find its first four terms.
Solution:

Given

The nth of a sequence is an = 2n+1.

an=2n+1

put n= 1,2,3,4, we get

a1 = 2×1+3 = 5

a2 = 2×2+3 = 7

a3 = 2×3+3 = 9

a4 = 2×4+3 = 11

∴ The first four terms of the sequence are 5,7,9,11.

Question 2. The nth term of a sequence is an=n2+5. Find its first terms.
Solution:

Given

The nth term of a sequence is an=n2+5

an=n2+5

put n=1,2,3, we get

a1 = 12 + 5 = 6

a2 = 22 + 5 = 9

a3 = 32 + 5 = 14

∴ The first four terms of the sequence are 6,9,14.

Question 3. Fibonacci sequence is defined as follows: a1 = a2 = 1 and an = an-2 + an-1 , where n > 2. Find third, fourth and fifth terms
Solution:

Given

Fibonacci sequence is defined as follows: a1 = a2 = 1 and an = an-2 + an-1 , where n > 2.

a1 = a1 = 1

⇒ \(a_n=a_{n-2}+a_{n-1}, n>2\)

put n = 3, we get

⇒ \(a_3=a_1+a_2=1+1=2\)

put n = 4, we get

⇒ \(a_4=a_2+a_3=1+2=3\)

put it = 5, we get

⇒ \(a_5=a_3+a_4=2+3=5\)

Question 4. A sequence is defined as follows: a1 = 3, an = 2an-1 + 1, where n > 1. Find \(\frac{a_{n+1}}{a_n}\) for n=1,2,3.
Solution:

Given

A sequence is defined as follows: a1 = 3, an = 2an-1 + 1, where n > 1.

a1=3

⇒ \(a_n=2 a_{n-1}+1 \quad \text { where } n>1\)

put n= 2, we get

⇒ \(a_2=2 a_1+1=2 \times 3+1=7\)

put n = 3, we get

⇒ \(a_3=2 a_2+1=2 \times 7+1=15\)

put n = 4, we get

\(a_4=2 a_3+1=2 \times 15+1=31\)

Now, for n = 1

⇒ \(\frac{a_{n+1}}{a_n}=\frac{a_2}{a_1}=\frac{7}{3}\)

For n = 2,

⇒ \(\frac{a_{n+1}}{a_n}=\frac{a_3}{a_2}=\frac{15}{7}\)

For n = 3,

⇒ \(\frac{a_{n+1}}{a_n}=\frac{a_4}{a_3}=\frac{31}{15}\)

Arithmetic Progression

An arithmetic progression (A.P.) is a list of numbers in which each term is obtained by adding a fixed number to the preceding term, except the first term.

or

If the difference of any two consecutive terms of a progression is the same (constant), it is called arithmetic progression.

For example: a, a + d, a + 2d, a + 3d, ….

Common Difference

The difference between two consecutive terms (i.c., any term — preceding term) of an arithmetic progression is called a common difference.

It is denoted by ‘d’

∴ d = a2-a1 = a3-a2= … = an-an-1.

General Term Of Arithmetic Progression

Let the first term and common difference of an arithmetic progression be ‘a’ and ‘b’ respectively.

∴ Arithmetic progression is a, a +d, a + 2d, a + 3d,…..

Here, first term = a = a+(1-1)d

Second term = a+d = a+(2-1)d

Third term = a+2d = a+(3-1)d
.
.
.
.

nth term = a+(n-1)d

∴ an=a+(n-1)d

nth term of a progression is called its general term.

nth Term From The End Of An Arithmetic Progression

Let the first term, common difference and last term of an arithmetic progression be a, d and / respectively.

Arithmetic progression is a, a + d, a + 2d,…….l – 2d, l-d,l

Here, the First term from the end -l = l – (1 -1)d

Second term from the end = l – d =l-(2-l)d

Third term from the end = l – 2d = l – (3 – 1 )d

.

.

.

nth term from the end =l-(n- 1 )d.

Arithmetic Progression Solved Examples

Question 1. For the following A.P., write the first term and common difference: 3, 1,-1,-3, …
Solution:

3, 1,-1,-3,…

Here, first term a = 3

Common difference = 1 – 3 = -2

Question 2. Write the first four terms of the A.P., when the first term V and the common difference ‘d’ are given as follows:

a = 10, d = 5

Solution:

a = 10, d = 5

a1 = 10

a2 =10 + 5=15

a3 =15 + 5 = 20

a4 = 20 + 5 = 25.

∴ First 4 terms are 10, 15, 20, 25.

Question 3. Find the 18th term of the A.P. 4, 7, 10, …
Solution:

Given

A.P. 4, 7, 10

Here, a = 4,

d = 7 – 4 = 10-7 = 3,

n = 18

∴ an = a + (n – 1 )d

a18 = 4 + (18 – 1) × 3 =4 + 51=55

∴ 18th term of the given A.P. = 55.

Question 4. What is the common difference of an A.P. in which a21-a7 = 84?
Solution:

Given

a21-a7 = 84

Let the first term of A.P. be ‘a’ and common difference be ‘d’.

Since, a21-a7 =84

∴ (a+20d) -(a-6d) = 84

⇒ 14d = 84

⇒ d= 6

Hence, the common difference is 6.

Question 5. Which term of the A.P. 3, 8, 13, 18,….is 88?
Solution:

Given

A.P. 3, 8, 13, 18,….is 88

Here, a = 3,

A = 8-3 = 13 -8 = 5

Let an = 88

⇒ 3 + (n – 1)5 = 88

⇒ 3 + 5K – 5 = 88

⇒ 5n – 2 = 88

⇒ 5n = 90

⇒ n= 18

The 18th term of the given A.P. is 88.

Question 6. Which term of the A.P. 90, 87, 84,…is zero?
Solution:

Given

90, 87, 84,…

Here, a = 90,

d = 87 – 90 = -3

Let an = 0

⇒ 90 + (n – 1) (-3) = 0

⇒ 90 – 3n + 3 = 0

⇒ -3n = -93

⇒ n= 31

∴ 31st term of the given A.P. is zero.

Question 7. If 2, a, b, c, d, e,f and 65 form an A.P., find the value of e.
Solution:

Given

2, a, b, c, d, e,f and 65 form an A.P.

Here, T1 = 2, T8 = 65, T6 = e

Let the common difference be D

T8 = 65 ⇒  T1 + (8 – 1)D = 65

⇒ 2+7D = 65  ⇒ D = 9

∴ e = T6 = T1 + (6 – 1 )D = 2 + 5(9) = 47

Hence, the value of e is 47.

Question 8. Which term of the A.P. \(10,9 \frac{1}{3}, 8 \frac{2}{3}\) ….. is the first negative term?
Solution:

Here, a = 10,

⇒ \(d=9 \frac{1}{3}-10=8 \frac{2}{3}-9 \frac{1}{3}=-\frac{2}{3}\)

Let an<0

⇒ \(10+(n-1)\left(-\frac{2}{3}\right)<0\)

⇒ \(\frac{30-2 n+2}{3}<0\)

⇒ 32-2n<0

⇒  32 < 2n

⇒ 2n>32

⇒ n>16

⇒ n = 17, 18, 19, (all are negative terms)

∴ First negative term = 17th term

Now, a17=a+(17-1) d

⇒ \(10+16\left(-\frac{2}{3}\right)=10-\frac{32}{3}=\frac{-2}{3}\)

Thus, 17th term which is \(-\frac{2}{3}\) is the first negative term.

Question 9. The 18th term of an A.P. exceeds its 10th term by 8. Find the common difference.
Solution:

The 18th term of an A.P. exceeds its 10th term by 8.

Let the first term and the common difference of the A.P. be ‘a’ and ‘d’ respectively.

Given that,

⇒ a18 = a10 + 8

⇒ a18-a10= 8

⇒ {a +(18-1)d}-{a + (10- 1)d} = 8

⇒ [a + 17d)-{a + 9d) = 8

⇒ a + 17d-a-9d = 8

⇒ 8d = 8

⇒ d = 1

∴ Common difference = 1.

Example 10. The 26th, 11th and last term of an A.P. are 0, 3 and \(-\frac{1}{5}\) respectively. Find the common difference and the number of terms.
Solution:

The 26th, 11th and last term of an A.P. are 0, 3 and \(-\frac{1}{5}\) respectively.

Let the first term, common difference and number of terms in the A.P. be a, d and n respectively.

Given that

⇒ a26 = 0

⇒ a+25d = 0 ……(1)

⇒ a = -25d

⇒ a11 = 3 a+10d = 3

⇒ -25d+10d = 3 [from (1)]

⇒ -15d = 3

⇒ d = \(-\frac{1}{5}\)

put d = \(-\frac{1}{5}\) in equation (1) we get

⇒ \(a=-25\left(-\frac{1}{5}\right)=5\) and \(a_n=-\frac{1}{5}\)

⇒ \(5+(n-1)\left(-\frac{1}{5}\right)=-\frac{1}{5}\)

⇒ \((n-1)\left(-\frac{1}{5}\right)=-\frac{1}{5}-5=-\frac{26}{5}\)

⇒ n- 1=26

⇒ n= 27

No. of terms in the A.P. = 27.

Question 11. If the 9th term of an A.P. is zero, prove that its 29th term is twice its 19th term.
Solution:

The 9th term of an A.P. is zero

Let the first term and the common difference of the A.P. be ‘a’ and ‘d’ respectively.

Given that,

a9 = 0

a + 8d = 0

a = -8d

Now, a29 = a+28d = -8d+28d

= 20d

T19 = a+18d = -8d + 18d

= 10d

⇒ 2T19 = 20d

∴ T29 = 2T19

Question 12. The nth term of a sequence is 3n + 5. Show that it is an A.P.
Solution:

Given

The nth term of a sequence is 3n + 5.

Here, an = 3n+5

an-1 = 3(n-1) +5

3n-3+5 = 3n+2

Now, an-an-1 = (3n+5)-(3n+2) = 3

which does not depend on n i.e., it is constant.

∴ Given sequence is in A.P.

Question 13. For what value of k will k + 9, 2k- 1 and 2k +7 are the consecutive terms of an A.P?
Solution:

Since k + 9, 2k- 1 and 2k + 7are in A.P.

So, there must be a common difference.

i.e., (2k- 1) – (k + 9) = (2k + 7) – (2k- 1)

⇒ k- 10 = 8

⇒ k= 18

The value of k= 18.

Question 14. Find how many integers between 200 and 500 are divisible by 8.
Solution:

Integers between 200 and 500 which are divisible by 8 are as follows :

∴ 208, 216, 224, 232, …, 496

This forms an A.P. whose first term = 208, common difference = 8 and last term = 496. Let there be n terms.

∴ an = 496

⇒ a+ (n -1)<d = 496

⇒ 208 + (n- 1)8 = 496

⇒ (n- 1)8 = 496-208 = 288

⇒ \(n-1=\frac{288}{8}\)

∴ n-1 = 36

∴ n = 37

Hence, 37 integers are between 200 to 500 which are divisible by 8.

Question 15. If m times the with term of an A.P. is equal to n times the mth term and m ≠ n, show that its (m + n)th term is zero.
Solution:

Let the first term and the common difference of the A.P. be ‘a and ‘d’ respectively.

Given that

m.am = n.an

⇒ \( m\{a+(m-1) d\}=n\{a+(n-1) d\}\)

⇒ \(a m+\left(m^2-m\right) d=a n+\left(n^2-n\right) d\)

⇒ \(a(m-n)+\left\{\left(m^2-n^2\right)-m+n\right\} d=0\)

⇒ \(a(m-n)+\{(m-n)(m+n)-1(m-n\} d=0\)

⇒ \(a(m-n)+(m-n)(m+n-1) d=0\)

⇒ \((m-n)\{a+(m+n-1) d\}=0\)

⇒ \(a+(m+n-1) d=0\) (…m=n)

⇒ \(a_{m+n}=0\)

∴ (m + n)th term of the given A.P. is zero.

Question 16. If the term of an A.P. is \(\frac{1}{n}\) and its nth term be \(\frac{1}{m}\) then shows that its (nm)th term is 1.
Solution:

Let the first term and the common difference of the A.P. be ‘a’ and ‘d’ respectively

∴ \(a_m=\frac{1}{n} \quad \Rightarrow \quad a+(m-1) d=\frac{1}{n}\) ….(1)

and \(a_n=\frac{1}{m} \quad \Rightarrow \quad a+(n-1) d=\frac{1}{m}\)

Subtract equation (2) from equation (1), we get

Arithmetic Progression If The mth Term Of AP

⇒ \((m-n) d=\frac{m-n}{n m}\)

⇒ \(d=\frac{1}{m n}\)

put \(d=\frac{1}{m n}\)

⇒ \( a+(m-1) \cdot \frac{1}{m n}=\frac{1}{n}\)

⇒ \(a+\frac{1}{n}-\frac{1}{m n}=\frac{1}{n}\)

⇒ \(a=\frac{1}{m n}\)

⇒ \(a_{m n}=a+(m n-1) d\)

⇒ \(\frac{1}{m n}+(m n-1) \cdot \frac{1}{m n}=\frac{1}{m n}+1-\frac{1}{m n}=1\)

∴ (mn)th term of the A.P. = 1

Question 17. Find the 7th term from the end of the A.P. : 3, 8, 13, 18,….98
Solution:

Here, l = 98,

d = 8-3 = 13-8 = 5,

n = 7

∴ 7th term from the end =l- (7 – 1)d

= 98 – 6 x 5

= 98 – 30

= 68

Alternative Method :

Write the A.P. in reverse order 98, …18, 13,8,3

Here, a = 98

d = 3-8 = 8- 13 = -5

a7 = a + (7 – 1)d

= 98 + 6 x (-5) = 98- 30 = 68

which is the 7th term from the end of the given A.P.

Sum Of n Terms Of An A.P.

Let the first term and common difference of an A.P. be ‘a’ and ‘d’          respectively. Let the A.P. contain ‘n’ terms and the last term be ‘l’.

∴ l = a + (n-1)d ….(1)

Now. a sum of n terms

Sn = a + (a + d) + + (1-d)+l …..(2)

In reverse order

Sn = l + (l + d) + + (a-d)+a …..(3)

Adding equations (2) and (3), we get

2Sn = (a + l) + (a + l) + …… + (a+l) + (a +l) (no.of terms = n)

2Sn = n(a+l)

⇒ \(S_n=\frac{n}{2}(a+l)\)

⇒ \(S_n=\frac{n}{2}[a+a+(n-1) d]\)

From eq.(1)

⇒ \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

Sum Of n Terms Solved Examples

Question 1. Find the sum of n terms of the series

⇒ \(\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+\ldots\)

Solution:

We are given the series

⇒ \(S=\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+\ldots \text { to } n \text { terms }\)

⇒ \((4+4+4+\ldots \text { to } n \text { terms })-\left(\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+\ldots \text { to } n \text { terms }\right)\)

⇒ \(4 n-\frac{1}{n}(1+2+3+\ldots \text { to } n \text { terms })\)

⇒ \(4 n-\frac{1}{n}\left[\frac{n}{2}\{2 \times 1+(n-1) \times 1\}\right] \quad( a=1, d=1)\)

⇒ \(4 n-\frac{1}{n}\left[\frac{n}{2}(n+1)\right]=4 n-\frac{n+1}{2}=\frac{8 n-n-1}{2}=\frac{1}{2}(7n-1)\)

Hence, the required sum of n terms is \(\frac{1}{2}(7 n-1) \text {.}\)

Question 2. Find the sum of the following series : 5 + (-41) + 9 +(-39) + 13 + (-37)+ 1 7 + … + (-5) + 81 + (-3)
Solution:

Let 5 = 5 + (-41) + 9 + (-39)+ 13 + (-37) + 17 + … + (-5) + 81 + (-3)

= (5 + 9+ 13 + 17 + … + 81) – (41 + 39 + 37 + … + 3)

= S1-S2

where, ,S1 = 5 + 9+ I3 + 17 + … + 81

It is an A.P. with a1 = 5,d = 9- 5 = 4

Let there be n terms.

∴ an = 81

⇒ a1 + (n- 1)d = 81

⇒ 5 + (n – 1)4 = 81

⇒ (n-1)4 = 76

⇒ n-1= 19

⇒ n = 20

∴ S1 = Sum of 20 terms with first term 5 and last term 81

⇒ \(\frac{20}{2}[5+81]=10 \times 86\)

=860

and S2 = 41 + 39 + 37 + … + 3

It is also an A.P. with a1 = 41, d = 39 – 41 = 2

Let there be n terms.

∴ an = 3

⇒ a1 + (n- 1)d = 3

⇒ 41 + (n – 1)(-2) = 3

⇒ (n- 1)(-2)=-38

⇒ n- 1 = 19

⇒ n=20

∴ S2 = Sum of 20 terms with first term 41 and last term 3

⇒ \(\frac{20}{2}[41+3]\)

⇒ 10×44 = 440

From equation (1),

S = S1 – S2

⇒ S = 860- 440 = 420

Question 3. The first term of an A.P. is 5, the last term is 45 and the sum of all its terms is 150.

Find the number of terms and the common difference of the A.P.

Solution:

Given

The first term of an A.P. is 5, the last term is 45 and the sum of all its terms is 150.

Let a number of terms = n, the first term is a = 5 and the last term is l = 45.

We have, Sn = 150

⇒ \(\frac{n}{2}[a+l]=150\)

⇒ \(\frac{n}{2}(5+45)=150\)

n = 6

Now, l = 45 i.e., nth term = 45

⇒ a + (n-1)d = 45

⇒ 5 + (6- 1)d = 45

⇒ 5d = 40

⇒ \(d=\frac{40}{5}\)

= 8

Hence, number of terms = 6 and common difference = 8

Question 4. Solve for x: 5+ 13 + 21 +……. +x = 2139
Solution:

Given

5+ 13 + 21 +……. +x = 2139

Here, a = 5,d= 13-5 = 8

Let Tn=x

∴ Sn = 2139

∴ \(\frac{n}{2}[2 \times 5+(n-1) 8]=2139\)

⇒ n(5 + 4n – 4)= 2139

⇒ 4n2 + n- 2139 = 0

⇒ 4n2 – 92n + 93n -2139 = 0

⇒ (n – 23) (4n + 93) = 0

∴ n = 23 or \(n=-\frac{93}{4}\)

since n > 0, we have n = 23

x = Tn = a + (n- 1 )d = 5 + (23 – 1 )(8)

x = 181

Question 5. How many terms of an A.P. -6, \(\frac{-11}{2}\),-5 are needed to give the sum -25? Explain the double answer.
Solution:

Here, a= -6, \(d=\frac{-11}{2}-(-6)=\frac{1}{2}\)

Let -25 be the sum of it terms of this A.P. (n ∈ N)

Using \(S_n=\frac{n}{2}[2 n+(n-1) d]\)

⇒ \(-25=\frac{n}{2}\left[2(-6)+(n-1)\left(\frac{1}{2}\right)\right]\)

⇒ \(-50=n\left(-12+\frac{n-1}{2}\right)\)

⇒ \(-50=n\left(\frac{n-25}{2}\right)\)

⇒ -100 = n2– 25n

⇒ n2-25n+ 100 = 0

⇒ (n – 5) (n – 20) = 0

∴ n = 5, 20

Both the values of n are natural and therefore, admissible.

Explanation of Double Answer:

S20 = \( -6-\frac{11}{2}-5-\frac{9}{2}-4-\frac{7}{2}-3-\frac{5}{2}-2-\frac{3}{2}-1-\frac{1}{2} +0+\frac{1}{2}+1+\frac{3}{2}+2+\frac{5}{2}+3+\frac{7}{2} \)

⇒ \(-6-\frac{11}{2}-5-\frac{9}{2}-4\)

⇒ S5

Question 6. Find the sum of all odd numbers lying between 100 and 200.
Solution:

The series formed by odd numbers lying between 100 and 200 is 101 + 103 + 105 + … + 199

Here, a = 101,

d= 103- 101 = 105- 103 = 2

Let, an= 199

⇒ 101 + (n- 1)2= 199

⇒ (n- 1)2 = 98

⇒ n-1 = 49

⇒ n = 50

Now, \(S_{50}=\frac{50}{2}(101+199)\)

=7500

The sum of all odd numbers lying between 100 and 200 =7500

Question 7. If an = 3 – 4n, show that a1,a2,a3, … form an A.P. Also find S20.
Solution:

Given

an = 3 – 4n

an = 3 – 4n

an-1 = 3-4(n-1)

= 3-4n+4=7-4n

∴ an-an-1 = (3-4n)-(7-4n) = -4n

Which does not depend on ‘n’ i.e., the difference of two consecutive terms is constant.

∴ Given sequence is in A.P.

Now, a1 = a = 3 – 4(1 ) = -1 ,

d = -4

∴ \( S_{20}=\frac{20}{2}[2 a+(20-1) d]\)

⇒ \(=10[2(-1)+19(-4)]\)

=-780

Question 8. If the sum of the first 6 terms of an A.P. is 36 and that of the first 1 6 terms is 256, find the sum of the first 10 terms.
Solution:

Given

The sum of the first 6 terms of an A.P. is 36 and that of the first 1 6 terms is 256

Let the first term and common difference of A.P. be ‘a’ and ‘d’ respectively.

S6 = 36

⇒ \(\frac{6}{2}[2 a+(6-1) d]=36\)

2a + 5d= 12 ……(1)

5,6 = 256

⇒ \(\frac{16}{2}[2 a+(16-1) d]=256\)

2a + 15d = 32 …..(2)

Subtract eq. (1) from eq. (2), we get

⇒ \(\begin{array}{r}
2 a+15 d=32 \\
2 a+5 d=12 \\
-\quad-\quad- \\
\hline 10 d=20
\end{array}\)

d = 2

put d = 2 in eq. (1), we get

2a + 5(2) = 12

⇒ 2a = 2

⇒ a = 1

Now, the sum of first 10 terms 510 = \(\frac{10}{2}[2 a+(10-1) d]=5[2(1)+9(2)]\)

= 100

The sum of first 10 terms = 100

Question 9. The sum of the first V terms of an A.P. whose first term is 8 and the common difference is 20, is equal to the sum of the first 2, i terms of another A.P. whose first term is -30 and the common difference is 8. Find ‘n’.
Solution:

Given

The sum of the first V terms of an A.P. whose first term is 8 and the common difference is 20, is equal to the sum of the first 2, i terms of another A.P. whose first term is -30 and the common difference is 8.

For first A.P., a = 8, d = 20

∴ \( S_n=\frac{n}{2}[2(8)+(n-1)(20)]\)

⇒ \(n(8+10 n-10)=10 n^2-2 n \)

For second A.P., a = -30, d = 8

⇒ \(S_{2 n}=\frac{2 n}{2}[2(-30)+(2 n-1)(8)]\)

⇒ \(n(-60+16 n-8)=16 n^2-68n\)

Question 10. If the pth term of an A.P. is x and the qth term is y. Show that the sum of first (p + q) terms is \(\frac{p+q}{2}\left\{x+y+\frac{x-y}{p-q}\right\}\)
Solution:

Given

The pth term of an A.P. is x and the qth term is y.

Let first term = n and common difference = d

Now, Tp = x a + (p- 1)d = x …(1)

and Tq + a(q- 1)d =y …(2)

Subtracting eq. (2) from eq. (1), we get

(p – q)d = x- y

⇒ \(d=\frac{x-y}{p-q}\)

Now, if you put the value of d in eq. (1 ) or (2), and find a then it will be a very tedious job. So, don’t find ‘a we will simplify especially:

⇒ \(S_{p+q}=\frac{p+q}{2}[2 a+(p+q-1) d]\)

Bifurcate the terms inside the bracket as

⇒ \(S_{p+q}=\frac{p+q}{2}[\{a+(p-1) d\}+\{a+(q-1) d\}+d]\)

⇒ \(S_{p+q}=\frac{p+q}{2}\left[x+y+\frac{x-y}{p-q}\right]\) [From (1) ,(2) and (3)]

Question 11. The sum of the first six terms of an A.P. is 42. The ratio of its 10th term to 30th term is 1 : 3. Calculate the first term and 1 3th term of the A.P.
Solution:

Given

The sum of the first six terms of an A.P. is 42. The ratio of its 10th term to 30th term is 1 : 3.

Let the first term and common difference of A.P. be V and ‘d’ respectively.

Given that,

⇒ \(\frac{a_{10}}{a_{30}}=\frac{1}{3}\)

⇒ \(\frac{a+9 d}{a+29 d}=\frac{1}{3}\)

⇒ 3a + 27d = a + 29d

⇒ 2a = 2d

⇒ a=d

and, S6 = 42

⇒ \(\frac{6}{2}(2 a+5 d)=42\)

3(2 d+5 d)=42

d = 2

a = 2

a13 = a+12d

= 2 + 12 × 2

= 26

The first term and 13th term of the A.P is 2 and 26.

Question 12. The ratio between the sum of fird=st n terms of two A.P.’s are in the ratio (7n-5): (5n+17). Find the ratio of their 10th terms.
Solution:

Given

The ratio between the sum of fird=st n terms of two A.P.’s are in the ratio (7n-5): (5n+17).

Let a1 and d1, be the first term and common difference of first A.P and let a2 and d2 be the first term and common difference of other A.P.

∴ \( \frac{\frac{n}{2}\left[2 a_1+(n-1) d_1\right]}{\frac{n}{2}\left[2 a_2+(n-1) d_2\right]}=\frac{7 n-5}{5 n+17}\)

⇒ \(\frac{2 a_1+(n-1) d_1}{2 a_2+(n-1) d_2} =\frac{7 n-5}{5 n+17}\)

⇒ \(\frac{a_1+\left(\frac{n-1}{2}\right) d_1}{a_2+\left(\frac{n-1}{2}\right) d_2}=\frac{7 n-5}{5 n+17}\)

Replace \(\frac{n-1}{2}\) by 9

⇒ \(\frac{n-1}{2}\)

⇒ n = 19

∴ From eq. (1)

⇒ \(\frac{a_1+9 d_1}{a_2+9 d_2}=\frac{7(19)-5}{5(19)+17}\)

⇒ \(\frac{T_{10}}{T_{10}^*}=\frac{128}{112}=\frac{8}{7}\)

∴ \(T_{10}: T_{10}^*=8: 7\)

Question 13. If the sum of the first ‘m’ terms of an A.P. be V and the sum of its first ‘n’ terms is ‘m’. then show that the sum of its first (m + n) terms is -(m +n).
Solution:

Given

The sum of the first ‘m’ terms of an A.P. be V and the sum of its first ‘n’ terms is ‘m’.

Let the first term and common difference of the A.P. be ‘a’ and ‘d’ respectively.

Given that,

Sm = n

⇒ \(\frac{m}{2}[2 a+(m-1) d] =n \)

⇒ \(2 a m+\left(m^2-m\right) d =2 n\) ….(1)

and Sn =m

⇒ \(\frac{n}{2}[2 a+(n-1) d] =m\)

⇒ \(2 a n+\left(n^2-n\right)d =2 m\) …..(2)

Subtract eq. (2) from eq. (1), we get

⇒ 2a(m – n) + {(m2 – m2) – (m – n))d = 2(n – m)

⇒ 2a(m – n) + {(m – n) (m + n) – (m – n)}d = -2(m – n)

⇒ 2d(m – n) + (m – n) (m + n – 1)d = -2(m – n)

⇒ 2a + (m + n – l)d = -2 …..(3)

Now, \(S_{m+n}=\frac{m+n}{2}\{2 a+(m+n-1) d\}\)

⇒ \(=\frac{m+n}{2}(-2)=-(m+n)\)

Hence proved.

Question 14. If the sum of first n, 2n and 3n terms of an A.P. be S1, S2 and S3 respectively, then prove that: S3 = 3(S2 – S1)
Solution:

Given

The sum of first n, 2n and 3n terms of an A.P. be S1, S2 and S3 respectively,

Let the first term and common difference of the A.P. be V and ld’ respectively.

S1 = sum of first ‘n’ terms

⇒ \(S_1=\frac{n}{2}[2 a+(n-1) d]\) ….(1)

S2 = sum of first ‘2n terms

⇒ \(\frac{2 n}{2}[2 a+(2 n-1) d]\)

⇒ \(S_2=\frac{n}{2}[4 a+(4 n-2) d]\) ….(2)

and S3 = sum of first ‘3n’ terms

⇒ \(S_3=\frac{3 n}{2}[2 a+(3 n-1) d]\) …..(3)

⇒ \(S_2-S_1=\frac{n}{2}[4 a+(4 n-2) d]-\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(\frac{n}{2}[4 a+(4 n-2) d-2 a-(n-1) d]=\frac{n}{2}[2 a+(3 n-1) d]\)

⇒ \(3\left(S_2-S_1\right)=\frac{3 n}{2}[2 a+(3 n-1) d]=S_3\)

Hence proved

Question 15. If the ratio of the sum of the first m and n terms of an A.P. is m2:n2, show that the ratio of its mth and nth terms is (2m – 1) : (2n – 1).
Solution:

Given

The ratio of the sum of the first m and n terms of an A.P. is m2:n2

⇒ \(\frac{S_m}{S_n}=\frac{\frac{m}{2}[2 a+(m-1) d]}{\frac{n}{2}[2 a+(n-1) d]}=\frac{m^2}{n^2} \)

⇒ \(\frac{2 a+(m-1) d}{2 a+(n-1) d}=\frac{m}{n}\)

⇒ \(\frac{a+\left(\frac{m-1}{2}\right) d}{a+\left(\frac{n-1}{2}\right) d}=\frac{m}{n} \ldots\)

Caution

Some students write as :

put \(\frac{m-1}{2}=m-1 \quad \Rightarrow \quad m-1=2 m-2\)

m-2m = -2+1

m = 1

which is wrong.

Here has mixed up the m’s of \(\frac{m-1}{2} \text { and } m-1\)

We want, \(\frac{T_m}{T_n} \text { i.e., } \frac{a+(m-1) d}{a+(n-1) d}\)

So, we replace \(\frac{m-1}{2}\) as m-1

i.e., m-1 as 2(m – 1)

⇒ m – 1 as 2m – 2

⇒ m as 2m -2+1

i.e., replace m by 2m – 1

Similarly, replace n by 2n – 1 in eq. (1), we get

∴ \( \frac{a+(m-1) d}{a+(\dot{n}-1) d}=\frac{2 m-1}{2 n-1}\)

i.e., \(\frac{T_m}{T_n}=\frac{2 m-1}{2 n-1}\)

Question 16. An A.P. consists of 37 terms. The sum of the three middlemost terms is 225 and the sum of the last three terms is 429. Find the A.P.
Solution:

Given

An A.P. consists of 37 terms. The sum of the three middlemost terms is 225 and the sum of the last three terms is 429.

Total number of terms (n) = 37, which is odd

∴ Middle term = \(\frac{37+1}{2}\) th term = 19th term

So, 3 middle most terms are 18th, 19th and 20th

∴ T18 +T19 + = 225

⇒ a + 17d + a+ 18d + a+ 19d- 225

⇒ 3a + 54d = 225

⇒ a=18d = 75 ……(1)

Also, a sum of the last 3 terms = 429

T35 + T36 + T37 = 429

⇒ a + 34d + a + 35d + a + 36d = 429

⇒ 3a + 105d = 429

⇒ a + 35d = 143 ……(2)

Solving, (1) and (2), we get d- 4 and a = 3

∴ Required A.P. is a, a + d, a + 2d, a + 3d,…

i.e., 3, 3 + 4, 3 + 2(4), 3 + 3(4)

i.e., 3, 7, 11, 15, …

The A.p = 3, 7, 11, 15, …

Question 17. 25 trees are planted in a straight line 5 metres apart from each other. To water them the gardner must bring water for each tree separately from a well 10 metres from the first tree in line with trees. Find the distance he will move in order to water all the trees beginning with the first if he starts from the well.
Solution:

Arithmetic Progression 25 Trees Are Planted In A Straight Line 5 Metre Apart From Each Other

Distance Covered by Gardner from well to well :

⇒ \(\underbrace{10+10}_{\text {well-1-well }}+\underbrace{(10+5)+(10+5)}_{\text {well-2-well }}+\underbrace{(10+2 \times 5)+(10+2 \times 5)}_{\text {well-3-well }}\)

⇒ \(+\underbrace{(10+23 \times 5)+(10+23 \times 5)}_{\text {well-24-well }}+\underbrace{(10+24 \times 5)}_{\text {well to } 25 \text { th tree }}\)

2[10 + (10 + 5) + (10 + 2×5) + ……+(10 + 23×5)] + (10 + 24×5)

⇒ \(2 \underbrace{[10+15+20+\ldots+125]}_{24 \text { terms }}+(10+120)\)

⇒ \(2 \times \frac{24}{2}[10+125]+130=24 \times 135+130=3240+130=3370 \mathrm{~m}\)

Question 18. A child puts one five-rupee coin of her savings in the piggy bank on the first day. She increases her savings by one five-rupee coin daily. If the biggy bank can hold 1 90 coins of five-rupees in all, find the number of days she can continue to put the five-rupee coins into it and find the total money she saved. Write your views on the habit of saving.
Solution:

Child’s savings day-wise are,

Arithmetic Progression Childs Savings Day Wise Are

We can have at most 190 coins

i.e., 1+2 + 3+ 4 + 5 + …to n terms =190

⇒ \(\frac{n}{2}[2 \times 1+(n-1) 1]=190\)

⇒ n(n + 1) = 380

⇒ n2 +n- 380 = 0

⇒ (n + 20) (n- 19) = 0

⇒ n = -20

or n = 19

But many coins cannot be negative

∴ n = 19

So, number of days =19

Total money she saved = 5+10+15+20 + … upto 19 terms

⇒ \(\frac{19}{2}[2 \times 5+(19-1) 5]\)

⇒ \(\frac{19}{2}[100]=\frac{1900}{2}\)

=950

So, number of days = 19

and total money she saved = ₹950

Arithmetic Mean Of Two Numbers

If three numbers are in A.P., then the middle term is called the arithmetic mean of the remaining two numbers.

Let A be the arithmetic mean of a and b.

∴ a, A, b are in A.P.

⇒ A – a = b – A

⇒ 2A = a +b

⇒ A= \(\frac{a+b}{2}\)

Selection Of Continuous Terms Of A.P.

  1. Three consecutive terms in A.P. a— d, a, a + d (common difference is d)
  2. Four consecutive terms in A.P. a — 3d, a — d, a+ d, a + 3d (common difference is 2d)
  3. Five consecutive terms in A.P. a — 2d, a — d, a, a + d, a + 2d (common difference is d)

Selection Of Continuous Terms Of A.P. Solved Examples

Question 1. Find the value of x for which (x + 2), 2x, (2x + 3) are three consecutive terms of A.P. 
Solution:

Given

(x + 2), 2x, (2x + 3) are three consecutive terms of A.P.

∴ 2x = \(\frac{(x+2)+(2 x+3)}{2}\)

⇒ 4x = 3x + 5

x = 5

The value of x = 5.

Question 2. Split 207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623.
Solution:

Let three parts be a- d, a, a + d.

∴ a-d + a+ a+ d = 207

⇒ 3a = 207

⇒ a = 69

and (a-d)-a = 4623

⇒ (69-d)69 = 4623

⇒ 69 – d = 67

⇒ d = 2

∴ a-d = 69 – 2 = 67

a= 69

a+d = 692 = 71

⇒ three parts are 67,69,71

Question 3. The angles of a triangle are in A.P. The greatest angle is twice the least. Find all the angles of the triangle.
Solution:

Given

The angles of a triangle are in A.P. The greatest angle is twice the least.

Let angles of the triangle be a – d, a, a + d

∴ a – d+a+a + d = 180°

⇒ 3a = 1 80°

⇒ a = 60°

and a + d = 2(a – d)

⇒ 60° + W = 2(60°- d)

⇒ 60° + d = 120° -2d

⇒ 3d = 60°

⇒ d = 20°

∴ a- d = 60°- 20° = 40°

a +d = 60° + 20° = 80°

∴ Three angles of triangle are 40°, 60°, 80°

Question 4. The angles of a quadrilateral are in A.P. and their common difference is 10°. Find the angles.
Solution:

Given

The angles of a quadrilateral are in A.P. and their common difference is 10°.

Let the angles of the quadrilateral be

a, a + 10°, a + 20°, a + 30°   (∵ common difference is 10°)

∴ a + (a + 10°) + (a + 20°) + (a + 30°) = 360°

⇒ 4a + 60° = 360°

⇒ 4a = 300°

⇒ a = 75°

∴ a+ 10°= 75°+ 10° = 85°

a + 20° = 75° + 20° = 95°

a + 30° = 75° + 30°= 105°

Hence, the angles are 75°, 85°, 95°, 105°.

Alternative Method :

Let the four angles of a quadrilateral are

a – 3d, a – d, a + d and a + 3d

∴ Here the common difference is 2d (remember)

∴ (a – 3d) + (a-d) + (a + d) + (a + 3d) = 360°

4a = 360°

⇒ a = 90°

the common difference is given to be 10°

i.e., 2d = 10°

⇒ d = 5°

Four angles are a – 3d = 90°- 3(5°) = 75°,

a – d 90°- 5° = 85°,

a + d = 90° + 5° = 95°,

a + 3d = 90° + 3(5°) = 105°

Question 5. There are m arithmetic means between 5 and -16 such that the ratio of the 7th mean to the (m -7)th mean is 1: 4. Find the value of m.
Solution:

Given

There are m arithmetic means between 5 and -16 such that the ratio of the 7th mean to the (m -7)th mean is 1: 4

Let A1, A2,…….Am be m arithmetic means between 5 and -16.

∴ 5,A1, A2, A3,…..Am, -16 are in A.P.

∴ \(T_{m+2}=-16\)

⇒ \(5+(m+1) d=-16\)

⇒ \(d=\frac{-21}{m+1}\)

⇒ \(\frac{A_7}{A_{m-7}}=\frac{1}{4}\)

⇒ \(\frac{T_8}{T_{m-6}}=\frac{1}{4}\)

⇒\(\frac{5+7 d}{5+(m-7) d}=\frac{1}{4}\)

⇒ \(\frac{5+7\left(\frac{-21}{m+1}\right)}{5+(m-7)\left(\frac{-21}{m+1}\right)} \frac{1}{4}\)

⇒ \(20-\frac{588}{m+1}=5-\frac{21(m-7)}{m+7}\)

⇒ 20m + 20- 588 = 5m +5- 21m + 147

⇒ 36m = 720

⇒ m = 20

The value of m = 20.

Question 6. The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.
Solution:

Given

The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number.

Let the 3 digits in A.P. at units, tens and hundredth places be a – d, a and a + d.

According to the first condition,

(a -d) +a + (a +d) = 15

⇒ 3a = 15

a = 5 …(1)

The number is (a – d) + 10a + 100 (a + d)….(1)

i.e., 111+ 99d …(2)

The number; on reversing the digits is

(a + d) + 1 0a + 100(A — d) i.e., 111+ 99d

∴ According to the 2nd condition,

111a- 99d = (11la + 99d) – 594

⇒ 198d = 594

⇒ d = 3 …(3)

∴ Required number = 111a + 99d [from (2)]

= 111(5) + 99(3) [from (1) and (3)]

= 555 + 297

= 852

Required number = 852

Arithmetic Progression Exercise 5.1

Question 1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

  1. File taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional km.
  2. The amount of air present in a cylinder when a vacuum pump removes — the
    air remaining in the cylinder at a time.
  3. The cost of digging a well after every metre of digging, when it costs ₹150 for the first metre rises by ₹50 for each subsequent metre.
  4. The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8% per annum.

Solution :

1. Fare of first kilometre = ₹15 

Fare of 2 kilolnetre = ₹(15 + 8) = ₹23

Fare of 3 kilometre = ₹(23 + 8) = ₹31

Fare of 4 kilometre = ₹(3 1 + 8) = ₹39

Now, a1 = 15, a2 = 23, a3 = 31, a4 = 39

a2 -a1 = 23 – 15 = 8

a3-a2 = 31 -23 = 8

a4 – a3 = 39 – 31 = 8

∴ The difference between two consecutive terms is constant.

∴ The taxi fare after each kilometre is in A.P

2. Let the initial volume of air in the cylinder = V

In the first pump,

air remove = \(\frac{V}{4}\)

Remaining air = \(V-\frac{V}{4}=\frac{3 V}{4}\)

In the second pump,

air remove = \(\frac{1}{4} \times \frac{3 V}{4}=\frac{3 V}{16}\)

Remaining air = \(\frac{3 V}{4}-\frac{3 V}{16}=\frac{9 V}{16}\)

Now, \(a_1=V, a_2=\frac{3 V}{4}, a_3=\frac{9 V}{16}\)

∴ \(a_2-a_1=\frac{3 V}{4}-V=-\frac{V}{4}\)

and \(a_3-a_2=\frac{9 V}{16}-\frac{3 V}{4}=-\frac{3 V}{16}\)

∵ a2– a1 ≠ a3– a2

∴ The volumes of air are not in A.P.

3. The cost of digging of first metre = ₹150

The cost of digging of 2 metres = ₹.(150 + 50)

= ₹200

The cost of digging of 3 metres

= ₹ (150 + 50 + 50)

= ₹250

The cost of digging of 4 metres

= ₹(150 + 50 + 50 + 50)

= ₹300

Now, a1=₹150, a2 = 200,a3 = ₹250, a4 = ₹300

∴ a2-a1 = ₹200-₹150 = ₹50

a3-a2= ₹250 -₹200 = ₹50

a4-a3= ₹300- ₹250 = ₹50

∵  The difference between two consecutive terms is constant.

The costs of digging each metre are in A.P.

4. Principal P = ₹10,000;

rate of interest R = ₹8%

Amount after 1 year,

⇒ \(A_1=P\left(1+\frac{R}{100}\right)^1=10,000\left(1+\frac{8}{100}\right)^1\)

⇒ \(=10,000 \times \frac{108}{100}=₹ 10,800\)

Amount after 2 years,

⇒ \(A_2=P\left(1+\frac{R}{100}\right)^2=10,000\left(1+\frac{8}{100}\right)^2\)

⇒ \(=10,000 \times \frac{108}{100} \times \frac{108}{100}=₹ 11664\)

Amount after 3 years

⇒ \(A_3=P\left(1+\frac{R}{100}\right)^3=10,000\left(1+\frac{8}{100}\right)^3\)

⇒ \(=10,000 \times \frac{108}{100} \times \frac{108}{100} \times \frac{108}{100}\)

= ₹12597.12

Now, A2-A1 = 11664- 10800 = 864

A3-A2= 12597.12- 11664 = 933.12

The difference between two consecutive terms is not the same.

∴ Amounts are not in A.P.

Question 2. Write the first four terms of the A.P., when the first term a and the common difference d are given as follows :

  1. a= 10, d= 10
  2. a =- 2, d = 0
  3. a = 4,d = -3
  4. a=-1, d=\(\frac{1}{2}\)

Solution:

1. a = 10, d = 10

First term = a = 10

Second term =a + d= 10 + 10 = 20

Third term = a + 2d = 10 + 2 × 10 = 30

Fourth term = a + 3d = 1 0 + 3 × 1 0 = 40

∴ The first four terms of A.P. are 10, 20, 30, 40

2. a = -2, d = 0

First term = a = -2

Second term = a + d = -2 + 0 = -2

Third term = a + 2d = -2 + 2 × 0 = -2

Fourth term = a + 3d = -2 + 3 × 0=-2

∴ The first four terms of A.P. are -2, -2, -2, -2

3. a= 4, d = -3

First term = a = 4

Second term = a + d = 4 + (-3) = 1

Third term = a + 2d = 4 + 2(-3) = -2

Fourth term = a + 3d = 4 + 3(-3) =-5

∴ The first four terms of A.P are 4, 1, -2, -5

4. a=-1, d = \(\frac{1}{2}\)

First term =a = -1

Second term = a + d = \(-1+\frac{1}{2}=-\frac{1}{2}\)

Third term= a + 2d = \(-1+2 \times \frac{1}{2}=0\)

Fourth term= a + 3d = \(-1+3 \times \frac{1}{2}=\frac{1}{2}\)

∴ First four terms of A.P. are = \(-1,-\frac{1}{2}, 0, \frac{1}{2}\)

5. a =- 1.25, d = -0.25

First term = a = -1.25

Second term = a+d =-1.25 + (-0.25) = -1.50

Third term =a + 2d = -1.25 + 2(-0.25)

= -1.75

Fourth term = a + 3d =- 1.25 + 3(-0.25.)

= -2.00

∴ The first four terms of A.P. are

= -1.25, -1.50, -1.75, -2.00

Question 3. For the following A.Ps., write the first term and the common difference :

  1. 3, 1,-1, -3,…
  2. -5, -1,3, 7,…
  3. \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots\)
  4. 0.6, 17,2.8,3.9,…

Solution:

1. 3, 1, -1, -3,…

First term a = 3

Common difference d = 1-3 = (-1)- 1 = -2

2. -5,-1, 3, 7,…

First term a = -5

Common difference d = -1-(-5) = 3-(-1) = 4

3. \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots\)

First term a = \(\frac{1}{3}\)

Common difference d = \(\frac{5}{3}-\frac{1}{3}=\frac{9}{3}-\frac{5}{3}=\frac{4}{3}\)

4. 0.6, 17,2.8,3.9,…

First term a = 0.6

Common difference d= 1.7 – 0.6

= 2.8-17=1.1

Question 4. Which of the following are A.P.s? If they form an A.P., find the common difference d and write three more terms.

  1. \(2,4,8,16,\)
  2. \(2, \frac{5}{2}, 3, \frac{7}{2},\)
  3. \(-1.2,-3.2,-5.2,-7.2,\)
  4. \(-10,-6,-2,2,\)
  5. \(3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2},\)
  6. \(0.2,0.22,0.222,0.2222,\)
  7. \(0,-4,-8,-12,\)
  8. \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},\)
  9. \(1,3,9,27,\)
  10. \(a, 2 a, 3 a, 4 a,\)
  11. \(a, a^2, a^3, a^4,\)
  12. \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32},\)
  13. \(\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12},\)
  14. \(1^2, 3^2, 5^2, 7^2,\)
  15. \(1^2, 5^2, 7^2, 73,\)

Solution:

1. 2,4, 8, 16,…

Here, a1 = 2, = 4, = 8, = 16

a2– a1 = 4- 2 = 2

a3-a2 = 8- 4 = 4

∵ a2-a1 ≠ a3-a2

∴ Given sequence is not an A.P.

2. \(2, \frac{5}{2}, 3, \frac{7}{2},\)

Here, \(a_1=2, a_2=\frac{5}{2}, a_3=3, a_4=\frac{7}{2}\)

⇒ \(a_2-a_1=\frac{5}{2}-2=\frac{1}{2}\)

∴ \(a_3-a_2=3-\frac{5}{2}=\frac{1}{2}\)

⇒ \(a_4-a_3=\frac{7}{2}-3=\frac{1}{2}\)

∵ \(a_2-a_1=a_3-a_2=a_4-a_3=\frac{1}{2}\)

∴ Given sequence is an A.P. and d = \(\frac{1}{2}\)

Now, fifth term \(a_5=a_4+d=\frac{7}{2}+\frac{1}{2}=4\)

Sixth term \(a_6=a_5+d=4+\frac{1}{2}=\frac{9}{2}\)

Seventh term \(a_7=a_6+d=\frac{9}{2}+\frac{1}{2}=5\)

∴ Next three terms of A.P. = 4, \(\frac{9}{2}\),5

3. -1.2, -3.2, -5.2, -7.2, …

Here a1 = -1.2, a2 =-3.2, a3 = -5.2, a4 =-7.2

∴ a2-a1 =(-3.2) -(-1.2) =-3.2+ 1.2= -2

a3 -a2 = (-5-2) – (-3.2) =-5.2 + 3.2 =-2

a4 – a3 = (-7.2) – (-5.2) = -7.2 + 5.2 =- 2

∵ a2 – a1 = a3 – a2 = a4 – a3 = -2

∴ Given sequence is A.P. and d = -2

Now, fifth term a5= a4 + d = -7.2 + (-2) = -9.2

Sixth term a6 = a5 + d = -9.2 + (-2) = -1 1 .2

Seventh term a7 =a6+ d = -11.2 + (-2) =-13.2

∴ Next three terms of A.P. =- 9.2, -1 1.2, -13.2

4. -10, -6, -2,2,…

Here a1 = -10, a2 =-6, a3 =-2, a4 = 2

∴ a2-a1=(-6)-(-10)=-6+ 10 = 4

a3 – a2 = (-2) – (-6) =-2 + 6 = 4

a4 – a3 = 2-(-2) = 2 + 2 = 4

∵ a2-a1 = a3 – a2 = a4 – a3 = 4

∴ Given sequence is A.P. and d = 4

Now, fifth term a5 = a4 + d = 2 + 4 = 6

Sixth term a6 = a5 + d = 6 + 4 = 10

Seventh term a7 = + d = 1 0 + 4 = 14

∴ Next three-term of A.P. = 6, 10, 14

5. \(3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2},\)

Here, \(a_1=3, a_2=3+\sqrt{2}, a_3=3+2 \sqrt{2},\)

⇒ \(a_4=3+3 \sqrt{2}\)

⇒ \(a_2-a_1=(3+\sqrt{2})-3=\sqrt{2}\)

∴ \(a_3-a_2=(3+2 \sqrt{2})-(3+\sqrt{2})=\sqrt{2}\)

⇒ \(a_4-a_3=(3+3 \sqrt{2})-(3+2 \sqrt{2})=\sqrt{2}\)

∵ \(a_2-a_1=a_3-a_2=a_4-a_3=\sqrt{2}\)

∴ Given sequence is A.P. and d = \(\sqrt{2}\)

Now, fifth term \(a_5=a_4+d\)

⇒ \(3+3 \sqrt{2}+\sqrt{2}=3+4 \sqrt{2}\)

Sixth term \(a^6=a^5+d\)

⇒ \(3+4 \sqrt{2}+\sqrt{2}=3+5 \sqrt{2}\)

Seventh term \(a_7=a_6+d\)

⇒ \(3+5 \sqrt{2}+\sqrt{2}=3+6 \sqrt{2}\)

Next three terms of A.P.

⇒ \(3+4 \sqrt{2}, 3+5 \sqrt{2}, 3+6 \sqrt{2}\)

6. 0.2,0.22,0.222,0.2222, …

Here a1 = 0.2, a2 = 0.22, a3 = 0.222, a4 = 0.2222

∴ a2-a1=0.22-0.2 = 0.02

a3-a2 = 0.222 -0.22 = 0.002

∵ a2 – a1 = a3 – a2

∴ Given sequence is not an A.P.

7. 0, -4, -8, -12, …

Here, a1 = 0, a2 = -4, a3 = -8, a4 = -12

∴ a2 – a1 = -4 – 0 = – 4

a3-a2 =-8- (-4) =- 8 + 4 = -4

∵ a4 – a3 = -12 – (-8) =-12 + 8 =- 4

a2 – a1 = a3 – a2 = a4 – a3 = -4

∴ Given sequence is A.P. and d =-4

Now, fifth term a5 =a4 + d =-12 + (-4) =-16

Sixth term a6 = a5 + d = -16 + (-4) = -20

Seventh term a7 = a6 + d = -20 + (-4) = -24

∴ Next three terms of A.P. =-16, -20, -24

8. \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots\)

Here, \( a_1=a_2=a_3=a_4=-\frac{1}{2}\)

∴ \(a_2-a_1=\left(-\frac{1}{2}\right)-\left(-\frac{1}{2}\right)=0\)

⇒ \(a_3-a_2=\left(-\frac{1}{2}\right)-\left(-\frac{1}{2}\right)=0\)

⇒ \(a_4-a_3=\left(-\frac{1}{2}\right)-\left(-\frac{1}{2}\right)=0\)

∴ a2– 1 = a3 – a2 = a4 – a3 = 0

∴ Given sequence is A.P. and d = 0

Now, fifth term \(a_5=a_4+d=\frac{-1}{2}+0=-\frac{1}{2}\)

Sixth term \(a_6=a_5+d=\frac{-1}{2}+0=-\frac{1}{2}\)

Seventh term \(a_7=a_6+d=\frac{-1}{2}+0=-\frac{1}{2}\)

∴ Next three terms of A.P. = \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\)

9. 1,3, 9,27,…

Here, a1 = 1, a2 = 3, a3 = 9, a4 = 27

∴ a2-a1 = 3-1=2

a3– a2 = 9- 3 = 6

∵ a2 -a1 ≠ a3– a2

∴ Given sequence is not an A.P.

10. a, 2a, 3a, 4a

Here, a1 = a, a2 = 2a, a2 = 3a, a4 = 4a

∴ a2-a1 =2a – a = a

a3 – a2 = 3a – 2a =a

a4 – a3= 4a – 3a = a

a2– a1 = a3 – a2 =a4– a3 = a

∴ Given sequence is A.P. and d = a

Now, fifth term a5 = a4 + d = 4a + a = 5a

Sixth term a6 = a5 + d = 5a + a = 6a

Seventh term a7 = a6 + d = 6a + a = 7a

∴ Next three terms of A.P. = 5A, 6a, 7a

11. a, a2, a3, a4 = ,…

Here, a1 = a, a2 = a2, a3 = a3, a4 = a4

∴ a2-a1 = a2 – a

a3 – a2 = a3-a2

a2– a1 ≠ a3– a2

∴ Given sequence is not an A.P.

12. \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \cdots\)

Here, \(=\sqrt{2}, 2 \sqrt{2}, 3 \sqrt{2}, 4 \sqrt{2}, \cdots\)

∴ \(a_1=\sqrt{2}, a_2=2 \sqrt{2}, a_3=3 \sqrt{2}, a_4=4 \sqrt{2}\)

⇒ \(a_2-a_1=2 \sqrt{2}-\sqrt{2}=\sqrt{2}\)

⇒ \(a_3-a_2=3 \sqrt{2}-2 \sqrt{2}=\sqrt{2}\)

⇒ \(a_4-a_3=4 \sqrt{2}-3 \sqrt{2}=\sqrt{2}\)

⇒ \(a_2-a_1=a_3-a_2=a_4-a_3\)

∴ Given sequence is A.P. and d = \(\sqrt{2}\)

Now, fifth term \(a_5=a_4+d=4 \sqrt{2}+\sqrt{2}=5 \sqrt{2}\)

Sixth term \(a_6=a_5+d=5 \sqrt{2}+\sqrt{2}=6 \sqrt{2}\)

Seventh term \(a_7=a_6+d=6 \sqrt{2}+\sqrt{2}=7 \sqrt{2}\)

∴ Next three terms of A.P. = \(5 \sqrt{2}, 6 \sqrt{2}, 7 \sqrt{2}\)

13. \(\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots\)

Here, \(\quad a_1=\sqrt{3}, a_2=\sqrt{6}, a_3=\sqrt{9}, a_4=\sqrt{12}\)

⇒ \(a_2-a_1=\sqrt{6}-\sqrt{3}=\sqrt{3}(\sqrt{2}-1)\)

⇒ \(a_3-a_2=\sqrt{9}-\sqrt{6}=\sqrt{3}(\sqrt{3}-\sqrt{2})\)

⇒ \(a_2-a_1 \neq a_3-a_2\)

∴ Given sequence is not an A.P.

14. \(1^2, 3^2, 5^2, 7^2, \ldots\)

Here, \(\quad a_1=1^2, a_2=3^2, a_3=5^2, a_4=7^2\)

∴ \(a_2-a_1=3^2-1^2=9-1=8\)

⇒ \(a_3-a_2=5^2-3^2\)

⇒ \(=25-9=16\)

∵ \(a_2-a_1\neq a_3-a_2\)

∴ Given sequence is not an A.P.

15. \(1^2, 5^2, 7^2, 73, \ldots\)

Here, \(a_1=1^2, a_2=5^2, a_3=7^2, a_4=73\)

∴ \(a_2-a_1=5^2-1^2=25-1=24\)

⇒ \(a_3-a_2=7^2-5^2=49-25=24\)

⇒ \(a_4-a_3=73-7^2=73-49=24\)

∵ \(a_2-a_1=a_3-a_2=a_4-a_3=24\)

∴ Given sequence is not an A.P. and d = 24

Now, fifth term a5 = a4 + d = 73 + 24 = 97

Sixth term a6= a5+d = 97 + 24

= 121 = 112

Seventh term a7=a6 + d= 121 + 24 = 145

∴ Next three terms of A.P. = 97, 112, 145

Arithmetic Progression Exercise 5.2

Question 1. Fill in the blanks in the following table given the first term. d the common difference and an nth term of the A.P. :

Arithmetic Progression The Common Difference An And nth Term Of The AP

Solution:

1. Here, a = 7, d = 3. n= 8

∴ an – a + (n -1)d

= 7 + (8 – 1) 3

= 7 + 7 × 3=28

Therefore, an =28

2. a =-18, n = 10, an = 0

∴ a + (n – 1)d = an

⇒ -18 + (10- 1)d = 0

⇒ 9d = 18

⇒ d = 2

3. d =-3, n= 1 8, an =-5

∴ a + (n – 1 )d = an

⇒ a + (18 – 1) (-3) =-5

⇒ a-51 =- 5

⇒ a = -5 + 51 =46

4. a = -18.9, d = 2.5, an = 3.6

∴ a + (n -1)d = an

-18.9 + (n -1) × 2.5 = 3.6

⇒ (n- 1) × 2.5 =3.6+18.9 = 22.5

⇒ \(n-1=\frac{22.5}{2.5}=9\)

⇒ n = 9+1 = 10

5. a = 3.5, d = 0, n= 105

∴ an = a + (n -1 )d

= 3.5 + (105 – 1) × 0

= 3.5

Question 2. Choose the correct choice in the following and justify :

  1. 30th term of the A.P.: 10, 7, 4, …, is
    1. 97
    2. 77
    3. -77
    4. -87
  2. 11th term of the A.P.: -3, 2 …, is
    1. 28
    2. 22
    3. -38
    4. \(-48 \frac{1}{2}\)

Solution:

1. Given A.P.: 10,7,4…

Mere n= 10,d= 7- 10 = 4- 7 = -3, n =30

∴ an =a( n – 1 )d

⇒  a30 = 10 + (30 – 1) (-3) = 10 – 87 – -77

2. Given A.P. : \(-3,-\frac{1}{2}, 2, \ldots\)

Here a = \(a=-3, d=-\frac{1}{2}-(-3)=2-\left(-\frac{1}{2}\right)=\frac{5}{2}\)

n = 11

∴ an = a + (n – 1 )d

⇒ \(a_{11}=-3+(11-1) \frac{5}{2}=-3+25=22\)

Question 3. In the following A.Ps., find the missing terms in the boxes :

Arithmetic Progression In The Following APs Find The Missing Terms In The Boxes

Solution:

1. Here, first term a = 2

Third term = 26

⇒ a + (3-1)d = 26 2 + 2d = 26

⇒ 2d = 24

⇒ d= 12

∴ Second term = a + d = 2 + 12 = 14

∴ Term of the box = 14

2. Second term =13

⇒ a + (2-1)d = 13 (where a = first term and d = common difference)

⇒ a+d =13 ….(1)

and Fourth term = 3

⇒ a + (4 – 1 )d= 3 ⇒ a + 3d = 3 ….(2)

Subtracting equation (1) from equation (2),

Arithmetic Progression Fourth Term Is 3 Subtracting Equation 1 And 2

⇒ d = -5

Put the value of d in equation (1)

a + (-5) = 13 ⇒ a = 13+5

= 18

Third term a2 + d = 13 + (-5) = 8

∴ Term of the boxes = 18. 8 respectively.

3. Here, first term a = 5

Fourth term \(a_4=9 \frac{1}{2}\)

⇒ \(a+3 d=\frac{19}{2}\)

⇒ \(5+3 d=\frac{19}{2}\)

⇒ \(3d=\frac{19}{2}-5=\frac{9}{2} \Rightarrow d=\frac{3}{2}\)

Now, second term \(a_2=a+d=5+\frac{3}{2}=\frac{13}{2}\)

Third term \(a_3=a_2+d=\frac{13}{2}+\frac{3}{2}=\frac{16}{2}=8\)

∴ Term of the boxes = \(\frac{13}{2}, 8\)

4. Here, first term a = -4

Sixth term = 6

⇒ a + 5d = 6 ⇒ -4 + 5d = 6

⇒ 5d = 6+4= 10

⇒ d =2

∴ Second term a2 = a + d = -4 + 2 = -2

Third term a3 = a2 + d = -2 + 2 = 0

Fourth term a4 = a3+d = 0 + 2 = 2

Fifth term a5= a4 + d = 2 + 2 = A

∴ Term of the boxes = -2, 0, 2, 4 respectively.

5. Here, second term a2 = 38

⇒ a + d= 38

Sixth term a5 = -22

⇒ a + 5d = -22

Subtracting equation (1) from (2),

Arithmetic Progression Here Second Term A2 Is 38

d = -15

Put the value of d in equation (1),

a + (-15) =38

⇒ a = 38+ 15=53

Third term a3 = a2 + d = 38 + (-15) = 23

Fourth term a4 = a3+d = 23 + (-15) = 8

fifth term a5 = a4 + d = 8 + (-15) -7

∴ Term of the boxes = 53, 23, 8, -7 respectively.

Question 4. Which term of the A.P. : 3, 8, 13, 18,… is 78?
Solution:

Given A.P. : 3, 8, 13, 18,…

Here a = 3, d = 8- 3 = 13-8 = 5

Let an = 78 ⇒ a + (n – 1 )d = 78

⇒ 3 + (11 – 1)5 = 78

⇒ (n- 1)5 =78-3 = 75

⇒ \(n-1=\frac{75}{5}=15\)

⇒ n= 15+ 1 = 16

∴ The 16th term is 78.

Question 5. Find the number of terms in each of the following A.Ps. :

  1. 7, 13, 19,…, 205
  2. 18, \(15 \frac{1}{2}\), 13…..-47

Solution:

Given A.P.: 7, 13, 19,…, 205

a=7

d= 13-7= 19- 13 = 6

Let an = 205

⇒ a + (n- l)d = 205

⇒ 7 + (n – 1)6 = 205

⇒ (n- 1)6 = 205-7 = 198

⇒ \(n-1=\frac{198}{6}=33\)

⇒ n = 33 + 1 = 34

∴ Number of terms in given A.P. = 34

2. Given A.P.:18, \(15 \frac{1}{2}\), 13…..-47

Here, a= 18

⇒ \(d=15 \frac{1}{2}-18=13-15 \frac{1}{2}=-2 \frac{1}{2}=\frac{-5}{2}\)

Let \(a_n=-47\)

⇒ \(a+(n-1) d=-47\)

⇒ \(18+(n-1)\left(\frac{-5}{2}\right)=-47\)

⇒ \((n-1)\left(\frac{-5}{2}\right)=-47-18=-65\)

⇒ \(n-1=(-65)\left(-\frac{2}{5}\right)=26\)

⇒ n = 26 + 1 = 27

∴ Number of terms in given A.P. = 27

Question 6. Check whether -150 is a term of the A.P. 1 1,8, 5. 2….
Solution:

Given A.P: 1 1, 8, 5, 2,…

⇒ a= 11, d=8- 11 =5- 8 = -3

Let an = -150

⇒ a + (n- 1 )d = -150

⇒ 11 + (n – 1) (-3) = -150

⇒ 11 – 3n + 3 = -150

⇒ 14 + 150 = 3

⇒ \(n=\frac{164}{3}=54 \frac{2}{3}\)

∵ The value of n is not a whole number.

∴ -150 is not a term of a given A.P.

Question 7. Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.
Solution:

Given

In AP 11th term is 38 and the 16th term is 73

Let a be the first term and d the common difference of the A.P.

Now, a11 = 38

⇒ n +(11 – 1)d = 38

⇒ a+ 10 = 38 ….(1)

and a16 = 73

⇒ a+ (16- 1)d = 73

⇒ a+15d=73 …..(2)

Subtracting equation (1) from (2),

Arithmetic Progression Find The 31st Term Of An AP

Put the value of d in equation (1),

a+ 10 × 7 = 38

⇒ a=38- 70 =-32

Now, the 31st term of A.P.

a31 = n + (31 – 1)d

=- 32 + 30 × 7

= -32 + 210= 178

The 31st term of an A.P = 178.

Question 8. An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:

Given

An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106.

Let a be the first term and d the common difference of A.P.

∴ a3 = 12 ⇒ a + (3 -1)d = 12

⇒ a+2d = 12 ….(1)

and last term= 50th term = 106

⇒ a + (50-1)d = 106

⇒ a + 49 d = 106 …..(2)

Subtracting equation (1) from (2)

Arithmetic Progression An AP Consists Of 50 Terms Of Which 3rd Term Is 12 And The Last Term Is 106

⇒ d = 2

Put the value of d in equation (1),

a + 2 × 2 = 12

a = 12-4 = 8

Now, 29th term = a + (29 – 1)d = 8 + 28 × 2

= 8 + 56 = 64

The 29th term = 64

Question 9. If the 3rd and the 9th terms of an A.P. are 4 and- 8 respectively, which term of this A.P. is zero?
Solution:

Given

The 3rd and the 9th terms of an A.P. are 4 and- 8 respectively,

Let a be the first term and d the common difference of A.P.

a3 = 4 ⇒ a + (3-1)d = 4

a + 2d = 4…..(1)

a9 = -8 ⇒ a + (9- 1 )d = -8

a + 8d = -8 ……(2)

Subtracting equation (1) from (2),

Arithmetic Progression If The 3rd And The 9th Terms Of AP

⇒ d = -2

Put the value of d in equation (1),

⇒ a + 2(-2) = 4

⇒ a = 4 + 4 =8

Now, let an = 0 a + (n- 1)d =0

⇒ 8 + (n- l)(-2) =0

⇒ 8- 2n + 2 = 0

⇒ -2n =-10

⇒ n =5

∴ 5th term of the progression is zero.

Question 10. The 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.
Solution:

Given

The 17th term of an A.P. exceeds its 10th term by 7.

Let a be the first term and d the common difference of A.P.

∴ a17 =a10 + 7

⇒ a + (17- 1)d = a + (10- 1)d + 7

16d-9d = 7 ⇒ 7d =7

⇒ d=1

Common difference of progression = 1

Question 11. Which term of the A.P. : 3, 15, 27, 39, … will be 132 more than its 54th term?
Solution:

Given A.P. : 3, 15, 27, 39…

a = 3, d=15-3 = 27-15=12

∴ a54 = a+ (54 -1)d = 3 + 53 × 12

= 3 + 636 = 639

Let an =a54 + 132

⇒ a + (n – 1)d =639+ 132

⇒ 3 + (n -1)12 =771

⇒ (n- 1)12 =771 -3 = 768

⇒ \((n-1)=\frac{768}{12}=64\)

⇒ n = 64 + 1 = 65

∴ Required term = 65th term.

Question 12. Two A.P.s have the same common difference. The difference between their 100th terms is 1 00, what is the difference between their 1,000th terms?
Solution:

Given

Two A.P.s have the same common difference. The difference between their 100th terms is 1 00

Let the first term be a and the common difference be d of first A.P.

Let the first term be A and the common difference be D of the second A.P.

100th term of first progression = + (100 -1)d

= a + 99d

100th term of the second progression

=A + (100- 1)d

=A + 99d

Difference of 100th terms of two progression

= 100

⇒ (a + 99d) – (A + 99d) = 100

⇒ a + 99d-A-99d = 100

⇒ a-A = 100 ……(1)

Again, the 1000th term of the first progression

= a + (1000-1)d

= a + 999d

1000th term of the second progression

=A + (1000-1)d

=A + 999d

Difference of 1000th terms of two progressions

= (a + 999d) – (A + 999d)

= a + 999d-A-999d

= a – A = 100 [from equation (1)]

Question 13. How many three-digit numbers by 12. are divisible by 7?
Solution:

Three digit numbers: 100, 101 102, …, 999

Three-digit numbers divisible by 7: 105, 112, 119,…, 994

Here, a = 105, d= 1 12 – 105 = 1 19 – 1 12 = 7

Let an = 994

⇒ a + (n-1)d =994 ⇒ 105 + (n -1)7 = 994

⇒ (n- 1)7 =994- 105 = 889

⇒ \(n-1=\frac{889}{7}=127\)

⇒ n= 127+ 1 = 128

∴ Number of 3-digit numbers divisible by 7.

= 128

Question 14. How many multiples of 4 lie between 10 and 250?
Solution: The multiples of 4 between 10 and 250 are :

12, 16, 20, …,248

Here a=12,d= 16-12 = 20-16 = 4

an =248

⇒ a + (n- 1)d =248

⇒12 + (n- 1)4=248

⇒ (n- 1)4 =248- 12 = 236

⇒ n-1 = 59

⇒ n = 59 + 1 = 60

∴ Multiples of 4 between 10 and 250 = 60

Question 15. For what value of n, are the nth terms of two A.Ps?: 63,65,67,… and 3, 10, 17,… equal?
Solution:

First A.P.: 63, 65, 67…

Here a = 63

d = 65 – 63 = 67- 65 = 2

an = a + (n – 1 )d

= 63 + (n – 1)2 = 63 + 2n – 2

= 2n + 61

Second A.P. 3, 10, 17, …

Here A =3,D= 10-3= 17- 10 = 7

An =A + (n -1)D = 3 + (n – 1)7

= 3 + 7n-7 = 7n-4

According to the problem, an =An

⇒ 2n + 61 = 7n – 4

⇒ n -7n = -4 – 61

⇒ -5n = -65

⇒ n = 13

∴ The 13th terms of given progressions are equal.

Question 16. Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:

Let the first term be a and common difference d of the A.P.

According to the problem,

a7 – a5 =12

⇒ (a + 6d) – (a + 4d) = 12

⇒ a+6d-a-4d = 12

⇒ 2d = 12

⇒ d = 6

⇒ a3 = 16

⇒ a + (3-1)d = 16

⇒ a + 2 × 6 = a + (3-1)d = 16

⇒ a + 2 × 6 = 16

⇒ a = 16 – 12 = 4

Now A.P.: 4, 4 + 6, 4 + 2 × 6,…

= 4, 10, 16, …

Question 17. Find the 20th term from the last term of the A.P. : 3, 8, 13, …. 253.
Solution:

Given A.P. : 3, 8, 13,…, 253

Here, last term l = 253, = 8- 3 = 13-8 = 5

20th term from the end =l-(n-1)d

= 253 – (20- 1) × 5

= 253- 19 × 5 = 253-95 = 158

∴ 20th term from the end of the progression

= 158

The 20th term from the last term of the A.P = 158

Question 18. The sum of the 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the
Solution :

Given

The sum of the 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44.

Let the first term be a and common difference d of the A.P.

∴ a4 + a8 =24 ⇒ a + 3d + a + 7d = 24

⇒ 2a + 10d =24

⇒ a+2d = 12 …(1)

and a6 + a10 = 44 ⇒ a + 5d + a + 9d = 44

⇒ 2a+14d =44

⇒ a + 7d =22

Subtracting equation (1) from (2),

Arithmetic Progression Let First Term Be A And Common Difference D Of The AP Subtracting Equation 1 from 2

⇒ d = 5

Put the value of d in equation (1),

a + 5 × 5= 12

⇒ a= 12-25 =-13

∴ Second term = a + d = -13 + 5 =-8

Third term-a + 2d =-13+2×5 = -3

So, the first three terms of A.P. are -13, -8,-3.

Question 19. Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?
Solution:

Given

Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year.

Salary in first year = ₹5000 Salary in second year = ₹5000 + ₹200 = ₹5200 Salary in third year = ₹5200 + ₹200 = ₹5400 Progression formed from the salary of each year ₹5000, ₹5200, ₹5400, …

Here, a2-a1 = 5200 -5000 = 200

a3-a2= 5400 -5200 = 200

a2 – a1 = -a2

⇒ The above progression is an A.P.

a = ₹5000, d = ₹200

Let in the nth year, the salary becomes ₹7000

Let in the nth year, the salary becomes ₹7000.

∴ an =7000

⇒ a + (n-1)d =7000

⇒ 5000 + (n- 1)200 =7000

⇒ (n- 1)200 =7000-5000

⇒ (n- 1)200 =2000

⇒ \(n-1=\frac{2000}{200}=10\)

⇒ n= 10+ 1 = 11

∴ In the 11th year, the salary of Subba Rao will be ₹7000.

Question 20. Ramkali saved ₹5 in the first week of the year and then increased her weekly savings by ₹1.75. If in the nth week, her weekly savings become ₹20.75, find n.
Solution:

Given

Ramkali saved ₹5 in the first week of the year and then increased her weekly savings by ₹1.75. If in the nth week, her weekly savings become ₹20.75

Saving of the first week = ₹5

∵ ₹1.75 is increasing in the savings of every week. The saving of every week from an A.P, in which.

a = ₹5 and d = ₹1.75

Let the saving in the nth week = ₹20.75

⇒ a + (n – 1) d = 20.75

⇒ 5 + (w- 1) (1.75) = 20.75

⇒ (n-1) (1.75) =20.75 -5 = 15.75

⇒ \(n-1=\frac{15.75}{1.75}=9\)

⇒ n=9 + 1 = 10

n= 10

Arithmetic Progression Exercise 5.3

Question 1. Find the sum of the following A.Ps.:

  1. 2, 7, 12, …, to 10 terms.
  2. -37, -33, -29, …, to 12 terms
  3. 0.6, 1.7, 2.8, …, to 100 terms.
  4. \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10},\)…, to 11 terms.

Solution:

1. 2, 7, 12…..to 10th term

Here, a = 2,d = 7 -2= 12-7 = 5,n= 10

From the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{10}=\frac{10}{2}[2 \times 2+(10-1) \times 5]\)

= 5(4 + 45) = 5×49 = 245

2. -37, -33, -29,…, to 12 terms

Here a =-37

d = -33- (-37) =-29- (-33) = 4, n = 12

From the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{12}=\frac{12}{2}[2 \times(-37)+(12-1) \times 4]\)

= 6(-74 + 44) = 6 x (-30) = -180

3. 0.6, 1.7, 2.8,…, to 100 terms

Here a = 0.6, d = 1.7 – 0.6 = 2.8 – 1.7 = 1.1 n= 100

From the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{100}=\frac{100}{2}[2 \times 0.6+(100-1) \times 1.1]\)

= 50[1.2 + 108.9] = 50×110.1

= 5505

4. \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots, \text { to } 11 \text { terms }\)

Here, \(a=\frac{1}{15}, d=\frac{1}{12}-\frac{1}{15}=\frac{1}{10}-\frac{1}{12}=\frac{1}{60}, n=11\)

From the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \( S_n=\frac{11}{2}\left[2 \times \frac{1}{15}+(11-1) \times \frac{1}{60}\right]\)

⇒ \(=\frac{11}{2}\left[\frac{8+10}{60}\right]=\frac{11}{2} \times \frac{18}{60}=\frac{33}{20}\)

Question 2. Find the sums given below :

  1. 7 +\(10 \frac{1}{2}\) + 14+ … + 84
  2. 34 + 32 + 30 + … + 10
  3. -5 + (-8) + (-11) + …(-230)

Solution:

1. 7 +\(10 \frac{1}{2}\) + 14+ … + 84

Here a = 7, d = \(10 \frac{1}{2}\) = \(14-10 \frac{1}{2}\)

⇒ \(3 \frac{1}{2}=\frac{7}{2}\)

Let an = 84

⇒ a + (n-1)d = 84 => \(7+(n-1) \frac{7}{2}=84\)

⇒ \( (n-1) \frac{7}{2}=84-7=77\)

⇒ \(n-1=77 \times \frac{2}{7}=22\)

⇒ \(n=22+1=23\)

From the formula, \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \( S_{23}=\frac{23}{2}\left[2 \times 7+(23-1) \times \frac{7}{2}\right]\)

⇒ \(\frac{23}{2}[14+77]=\frac{23 \times 91}{2}\)

⇒ \(\frac{2093}{2}\)

⇒ \(1046 \frac{1}{2}\)

2. 34 + 32 + 30 + … + 10

Here a = 34, = 32- 34 = 30 – 32 = -2

Let an = 10 ⇒ a + (n – 1)d = 10

⇒ 34 + (n-1)(-2) =10

⇒ (n- 1) (-2) = 10-34

⇒ (n-1) (-2) =-24

⇒ n-1 =12

⇒ n=12+1 = 13

From the formula, \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{13}=\frac{13}{2}[2 \times 34+(13-1)(-2)]\)

⇒ \(\frac{13}{2}[68-24]\)

⇒ \(\frac{13}{2} \times 44=286 \quad\)

3. -5 + (_8) + (-11)… + (-230)

Here a =-5, d =-8- (-5) =-11 – (-8) =-3

Let an = -230 ⇒ a + (n-1)d = -230

⇒ -5 + (n – 1) (-3) = -230

⇒ (n- 1) (-3) =-230 + 5

⇒ (n – 1)(-3) = -225

⇒ n- 1 = 75

⇒ n = 75 + 1 = 76

From the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{76}=\frac{76}{2}[2 \times(-5)+(76-1)(-3)]\)

= 38(-10 – 225)

= 38 × (-235) = -8930

Question 3. In an A.P. :

  1. Given a = 5, d = 3, an = 50, find ii and Sn.
  2. Given a = 7, a13 = 35, find d and S13.
  3. Given a12 = 37, d = 3, find a and S12.
  4. Given a3 = 15, S10 = 125, findd and a10.
  5. Given d = 5, S9 = 75, find a and a9.
  6. Given a = 2, d = 8, Sn = 90, find n and an.
  7. Given a = 8, an = 62, Sn = 210, find n and d.
  8. Given an = 4, n = 2, Sn = -14, find n and a.
  9. Given a = 3, n = 8, S = 192, find d.
  10. Given l = 28, S = 144, and there are a total 9 terms. Find a

Solution:

1. a = 5,d = 3, an = 50

⇒ a + (n -1)d = 50

⇒ 5 + (n-1)3 =50

⇒  (n-1)3 = 45

⇒ a-1= 15

⇒ n = 16

Now, \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(\frac{16}{2}[2 \times 5+(16-1) \times 3]\)

⇒ \(8(10+45)=8 \times 55=440\)

n = 16 and Sn = 440

2. a = 7

a13= 35

⇒ a + (13 – 1)d = 35

⇒ 7 + 12d = 35

⇒ 12d =35-7 = 28

⇒ \(d=\frac{28}{12}=\frac{7}{3}\)

and, from \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{13}=\frac{13}{2}\left[2 \times 7+(13-1) \times \frac{7}{3}\right]\)

⇒ \(\frac{13}{2}[14+28]=\frac{13}{2} \times 42=273\)

∴ \(d=\frac{7}{3} \text { and } S_{13}=273 \quad\)

3. d = 3

a12 = 37

a+(12-1)d = 37

a + 11 × 3 = 37

a = 37 – 33 = 4

and, from \( S_n=\frac{n}{2}[2 n+(n-1) d]\)

⇒ \(S_{12}=\frac{12}{2}[2 \times 4+(12-1) \times 3]\)

⇒ \(6[8+33]=6 \times 41=246\)

∴ \(a=4, S_{12}=246\)

4. Let the first term be a and the common difference is d.

a3 = 15 ⇒ a + (3 -1)d= 15

⇒ a+2d = 15 …(1)

and S10= 125

⇒ \(\frac{10}{2}[2 a+(10-1) d]=125\)

⇒ 5[2a + 9d] = 125

⇒ 2a + 9d = 25 …(2)

Multiply equation (1) by 2 and subtracting from equation (2),

Arithmetic Progression The First Term Of AP Be A And Common Difference D Multiply Equation 1 by 2

d = -1

Put the value of d in equation (1)

⇒ a + 2 × – 1 = 15

⇒ a-2= 15

⇒ a = 15+2= 17

a10 = a+ (10-1)d

= 17 + 9(-1) = 17-9 = 8

∴ d = -1, a10 = 8

5. d = 5

and S9 = 75

⇒ \(\frac{9}{2}[2 a+(9-1) \times 5]=75\)

⇒ \( 2 a+40=\frac{75 \times 2}{9}\)

⇒ \(a +20=\frac{75}{9} \Rightarrow a+20=\frac{25}{3}\)

⇒ \(a=\frac{25}{3}-20=\frac{25-60}{3}\)

⇒ \(a=\frac{-35}{3}\)

and \(a_9=a+8 d=\frac{-35}{3}+8 \times 5=\frac{-35}{3}+40\)

⇒ \(\frac{-35+120}{3}=\frac{85}{3}\)

∴ \(a=\frac{-35}{3} \text { and } a_9=\frac{85}{3}\)

6. a = 2, d= 8

Sn = 90

⇒ \( \frac{n}{2}[2 a+(n-1) \cdot d]=90\)

⇒ \(\frac{n}{2}[2 \times 2+(n-1) \cdot 8]=90\)

⇒ \(\frac{4n}{2}[1+(n-1) \cdot 2]=90\)

⇒ \( n(2 n-1)=\frac{90 \times 2}{4}\)

⇒ \(2 n^2-n=45\)

⇒ \(2 n^2-n-45=0\)

⇒ \(2 n^2-10 n+9 n-45=0\)

⇒ \(2 n(n-5)+9(n-5)=0\)

⇒ \((n-5)(2 n+9)=0\)

⇒ \(n-5=0 \quad \text { or } 2 n+9=0\)

⇒ \(n =5 \quad \text { or } \quad n=-\frac{9}{2}\)

⇒ \(n =-\frac{9}{2} \text { is not possible.}\)

∴ n = 5

Now, an = a + (n-1)d

a5 = 2 + (5 – 1) × 8 = 2 + 32 = 34

n = 5 and an = 34

7. a = 8, an = 62

Sn= 210

⇒ \(\frac{n}{2}\left(a+a_n\right)=210\)

⇒ \(\frac{n}{2}(8+62)=210\)

⇒ \(n=\frac{210 \times 2}{70}=6\)

⇒ \(a_n=62\)

⇒ \(a+(n-1) d=62\)

⇒ 8 + (6-1)d= 62 => 5d = 62 – 8 = 54

⇒ \(d=\frac{54}{5}\)

∴ \(n=6, d=\frac{54}{5}\)

8. an = 4, d = 2 and Sn=-14

an = 4

⇒ a + (n -1)d = 4

⇒ a + (n – 1).2 = 4

⇒ a + 2n – 2 =4

⇒ a + 2n = 6

and Sn = -14 …(1)

⇒ \( \frac{n}{2}\left(a+a_n\right)=-14\)

⇒ \(\frac{n}{2}(a+4)=-14\)

⇒ \(\frac{n}{2}(6-2 n+4)=-14 \quad \text { from eqn. (1) }\)

⇒ \(\frac{n}{2}(10-2 n)=-14\)

⇒ \(n(5-n)=-14\)

⇒ \(5 n-n^2=-14\)

⇒ \(0=n^2-5 n-14\)

⇒ \(n^2-7 n+2 n-14=0\)

⇒ \( n(n-7)+2(n-7)=0\)

⇒ \((n-7)(n+2)=0\)

⇒ \(n-7=0 \text { or } n+2=0\)

⇒ \(n=7 \text { or } \quad n=-2\)

⇒ \(n=-2 \text { is not possible.}\)

∴ n=7

From equation (1)

a + 2 × 7 = 6

a = 5

a = 6 – 14= -8

∴ n = 7 and a = -8

9. \(a=3, n=8 \text { and } S=192\)

⇒ \(S=192\)

⇒ \(\frac{n}{2}[2 a+(n-1) d]=192\)

⇒ \(\frac{8}{2}[2 \times 3+(8-1) d]=192\)

⇒ \(4(6+7 d)=192\)

⇒ \(24+28 d=192\)

⇒ \(28 d=192-24=168\)

⇒ \(d=\frac{168}{28}=6\)

∴ d=6

10. l=28, S =144, n=9

S=144

⇒ \(\frac{n}{2}(a+l)=144 \Rightarrow \frac{9}{2}(a+28)=144\)

⇒ \(a+28=\frac{144 \times 2}{9}=32\)

⇒ a=32-28=4

∴ a=4

Question 4. How many terms of the A.P.: 9, 17, 25,… must be taken to give a sum of 636?
Solution:

Given A.P: 9. 17. 25….

Here a = 9,d= 17-9 = 25-17 = 8

Let Sn=636

⇒ \(\frac{n}{2}[2 a+(n-1)] d=636\)

⇒ \(\frac{n}{2}[2 \times 9+(n-1) \cdot 8]=636\).

⇒ \(n[9+(n-1) \cdot 4]=636\)

⇒ \(n(9+4 n-4)=636\)

⇒ \(n(4 n+5)=636\)

⇒ \(4 n^2+5 n-636=0\)

⇒ \(4 n^2+53 n-48 n-636=0\)

⇒ \( n(4 n+53)-12(4 n+53)=0\)

⇒ \((4 n+53)(n-12)=0\)

⇒ \(4 n+53=0 \text { or } n-12=0\)

⇒ \(n=-\frac{53}{4} \text { or } n=12\)

but \(n=-\frac{53}{4}\) is not possible.

∴ n = 12

Therefore, the number of terms = 12

Question 5. The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:

Here a = 5

Let the number of terms = n

⇒ \(a_n=45 \text { and } S_n=400\)

⇒ \(S_n=400 \Rightarrow \frac{n}{2}\left(a+a_n\right)=400\)

⇒ \(\frac{n}{2}(5+45)=400\)

⇒ \(n=\frac{400 \times 2}{50}=16\)

⇒ \(a_n=45\)

⇒ \(a+(n-1) d=45\)

⇒ \(5+(16-1) d=45\)

⇒ \(15 d=45-5=40\)

⇒ \(d=\frac{40}{15}=\frac{8}{3}\)

∴ \(n=16 \text { and } d=\frac{8}{3}\)

The number of terms and the common difference 16 and\(\frac{8}{3}\)

Question 6. The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:

Given

The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9

Let the number of terms = n

a= 17, d = 9

an = 350 ⇒ a + {n- l)d = 350

⇒ \( 17+(n-1) \cdot 9=350\)

⇒ \((n-1)9=350-17=333\)

⇒ \(n-1=\frac{333}{9}=37\)

⇒ \(n=37+1=38\)

Now, from \( S_n=\frac{n}{2}\left(a+a_n\right)\)

⇒ \(S_{38}=\frac{38}{2}(17+350)\)

⇒ \(19 \times 367=6973\)

∴ n=38 and Sn=6973

Question 7. Find the sum of the first 22 terms of an A.P. in which d = 7 and the 22nd term is 149.
Solution:

d =7

a22=149

⇒ \(a+(22-1) d=149\)

⇒ \(a+21 \times 7=149\)

⇒ \(a+147=149\)

⇒ \(a=149-147=2\)

⇒ \(S_n=\frac{n}{2}\left(a+a_n\right)\)

⇒ \(S_{22}=\frac{22}{2}\left(a+a_{22}\right)=11(2+149)\)

= 11 × 151 = 1661

The sum of the first 22 terms of an A.P. = 1661

Question 8. Find the sum of the first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.
Solution :

Given

Second and third terms of an A.P are 14 and 18 respectively

Let the first form of A.P. be a and the common difference be d.

Now, \(a_2=14 \quad \Rightarrow \quad a+d=14 \quad \ldots(1)\)

and \(a_3=18 \quad \Rightarrow a+2 d=18 \quad \ldots(2)\).

Subtracting equation (1) from (2)

Arithmetic Progression The First Term Of AP Be A And Common Difference D From Equation 1 And 2

Put the value of d in equation (l),

a+4 = 14

a = 14-4 = 10

Now, from the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{51}=\frac{51}{2}[2 \times 10+(51-1) \times 4]\)

⇒ \(\frac{51}{2}(20+200)\)

⇒ \(\frac{51}{2} \times 220=5610\)

∴ The sum of 51 terms = 5610

Question 9. If the sum of first 7 terms of an A.P is 49 and that of 1 7 terms is 289, find the sum of first n terms.
Solution:

Given

The sum of first 7 terms of an A.P is 49 and that of 1 7 terms is 289

Let the first term of A.P. be a and common difference be d.

S7 = 49

⇒ \( \frac{7}{2}[2 a+(7-1) d]=49 \Rightarrow \frac{1}{2}[2 a+6 d]=7\)

⇒ \(a+3 d=7\)

⇒ \(S_{17}=289\)

⇒ \(\frac{17}{2}[2 a+(17-1) d]=289\)

⇒ \(\frac{1}{2}[2 a+16 d]=17 \Rightarrow a+8 d=17 \ldots(2)\)

Subtracting equation (1) from (2),

Arithmetic Progression The First Term Of AP Be A And Common Difference D

d = 2

Put the value of d in equation (1),

a + 3×2=7

⇒ a + 6=7

⇒ a =7 -6=1

Now, \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(\frac{n}{2}[2 \times 1+(n-1) \cdot 2]\)

⇒\(n(1+n-1)=n^2\)

∴ The sum of n terms of A.P. = n2

Question 10. Show that a1, a2, an A.P. where a„ is defined as below :

  1. an = 3 + 4n
  2. an = 9 – 5n

Also, find the sum of the first 15 terms in each case.

Solution:

the nth term of the sequence,

an=3+4n

Put n = 1\(a_1=3+4 \times 1=3+4=7\)

Put n = 2\(a_2=3+4 \times 2=3+8=11\)

Put n = 3\(a_3=3+4 \times 3=3+12=15\)

Now, \(a_2-a_1=11-7=4\)

⇒ \(a_3-a_2=15-11=4\)

⇒ \( a_2-a_1=a_3-a_2=4\)

∴ The difference between two consecutive terms of the sequence is constant.

So, the sequence is A.P.

Now, d = 4 and a = 7

A sum of first 15 terms

⇒ \(frac{n}{2}[2 a+(n-1) d]\)

⇒ \(\frac{15}{2}[2 \times 7+(15-1) \times 4]\)

⇒ \(\frac{15}{2}[14+56]=\frac{15}{2} \times 70=525\)

2. \(a_n=9-5 n\)

Put n = 1, \(a_1=9-5 \times 1=9-5=4\)

Put n = 2, \(a_2=9-5 \times 2=9-10=-1\)

Put n = 3, \(a_3=9-5 \times 3=9-15=-6\)

Now, \(a_2-a_1=-1-4=-5\)

⇒ \(a_3-a_2=6-(-1)=-6+1=-5\)

⇒ \(a_2-a_1=a_3-a_2=-5\)

∴ The difference between two consecutive terms of the sequence is constant.

So, the sequence is A.P.

Now d =-5,a = 4

∴ The sum of the first 15 terms

⇒ \(\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(\frac{15}{2}[2 \times 4+(15-1)(-5)]\)

⇒ \(\frac{15}{2}(8-70)=\frac{15}{2} \times(-62)=-465\)

Question 11. If the sum of the first n terms of an A.P. is 4n – n2, what is the first term (that is S1)? What is the sum of the first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Solution:

Here Sn = 4n – n2

Put n = 1

⇒ \(S_1=4 \times 1-(1)^2=4-1=3\)

Put n = 2

⇒ \(S_2=4 \times 2-2^2=8-4=4\)

Second term \(a_2=S_2-S_1=4-3=1\)

Put n = 3

⇒ \(S_3=4 \times 3-3^2=12-9=3\)

⇒ \(a_3=S_3-S_2=3-4=-1\)

Put n = 9

⇒ \(S_9=4 \times 9-9^2=36-81=-45\)

Put n = 10

⇒ \( S_{10}=4 \times 10-10^2=40-100=-60\)

⇒ \(a_{10}=S_{10}-S_9=(-60)-(-45)\)

= -60+45 = -15

Replace n by (n – 1)

⇒ \(S_{n-1}=4(n-1)-(n-1)^2\)

⇒ \(4 n-4-\left(n^2-2 n+1\right)\)

⇒ \(4 n-4-n^2+2 n-1=6 n-n^2-5\)

⇒ \(a_n=S_n-S_{n-1}\)

⇒ \(\left(4 n-n^2\right)-\left(6 n-n^2-5\right)\)

⇒ \(4 n-n^2-6 n+n^2+5\)

⇒ 5-2 n

Question 12. Find the sum of the first 40 positive integers divisible by 6.
Solution:

The progression formed the positive integers divisible by 6

6, 12, 1 8, 24, … to 40 terms

Here a = 6, = 6, n = 40

From the formula, \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{40}=\frac{40}{2}[2 \times 6+(40-1) \times 6]\)

= 20(12 + 234) = 20 x 246 = 4920

The sum of 40 terms = 4920

Question 13. Find the sum of the first 15 multiples of 8.
Solution:

The first 15 multiples of 8 are 8, 16, 24, …, to 15 terms

Here, a = 8, d = 16 — 8 = 24 — 16 = 8, n= 15

From the formula \( S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{15}=\frac{15}{2}[2 \times 8+(15-1) \times 8]\)

⇒ \(\frac{15}{2} \times 8[2+14]\)

= 60×16 = 960

The sum of the first 15 multiples of 8 = 960

Question 14. Find the sum of the odd numbers between 0 and 50.
Solution:

Odd numbers between 0 and 50 are 1,3,5…..49.

Here, a= 1,d =3- 1 =5-3=2

Let, an = 49

⇒ 1+(n-1).2 = 49

⇒ 1 + 2n – 2 = 49

⇒ 2n – 2 = 49

⇒ 2n = 49 + 1 = 50

⇒ n = 25

Now, from the formula \(S_n=\frac{n}{2}\left(a+a_n\right)\)

⇒ \(S_{25}=\frac{25}{2}(1+49)\)

⇒ \(\frac{25}{2} \times 50=625\)

The sum of the odd numbers between 0 and 50 is 625.

Question 15. A contract on a construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc., the penalty for each succeeding day being? 50 more than for the preceding day. How much money does the contractor have to pay as a penalty if he has delayed the work by 30 days?
Solution:

Penalty for delay of first day = ₹200

Penalty for delay of second day = ₹250

Penalty for delay of third day = ₹300

This progression is an A.P.

Here a = 200, d = 250 – 200 = 300 – 250 = 50, n = 30

From the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \( S_{30}=\frac{30}{2}[2 \times 200+(30-1) \times 50]\)

=15(400+1450)

=15×1850=27750

The contractor will pay the penalty of ₹27750.

Question 16. A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.
Solution:

Let first prize = ₹a

Common difference d =- 20, n = 7

Given, S7 = 700

⇒ \(\frac{7}{2}[2 a+(7-1)(-20)]=700\)

⇒ \({[2 a-120]=200}\)

⇒ 2a =200+ 120

⇒ 2a =320

⇒ a = 160

∴ a2 =a + d= 160-20= 140

⇒ \(a_3=a_2+d=140-20=120\)

⇒ \(a_4=a_3+d=120-20=100\)

⇒ \(a_5=a_4+d=100-20=80\)

⇒ \(a_6=a_5+d=80-20=60\)

⇒ \(a_7=a_6+d=60-20=40\)

Prizes are ₹160, ₹140, ₹120, ₹100, ₹80, ₹ 60 and ₹40.

Question 17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, would be the same as the class, in which they are studying, There are three sections of each class. How many trees will be planted by the students?
Solution:

Arithmetic Progression There Are Three Sections Of Each Class.

The sequence so formed : 3, 6, 9, … is an A.P.

Here, a = 3,d = 6- 3 = 9- 6 = 3

and n= 12

Now, from the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{12}=\frac{12}{2}[2 \times 3+(12-1) \times 3]\)

= 6(6 + 33) = 6 x 39 = 234

Total trees planted = 234

Question 18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with the centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm,…. What is the total length of such a spiral made up of thirteen consecutive semicircles? ( Take \(\pi=\frac{22}{7}\))

Arithmetic Progression A spiral Is Made Up Of Successive Semicircles

Solution:

The radius of first semicircle r1 = 0.5 cm.

The radius of second semicircle r2 =1.0 cm

The radius of third semicircle r3 = 1.5 cm

.
.
.
.
.
.
.

This sequence is an A.P.

Here a = 0.5 and d = 1.0 – 0.5 = 0.5, n = 13

Now, the length of the spiral is made of 1 3 consecutive semicircles.

⇒ \(\pi r_1+\pi r_2+\pi r_3+\ldots \text { to } 13 \text { terms }\)

⇒ \(\pi\left[r_1+r_2+r_3+\ldots 13\right.\)

⇒ \(\frac{22}{7} \times \frac{13}{2}[2 a+(13-1) \cdot d]\)

⇒ \(\frac{143}{7}[2 \times 0.5+12 \times 0.5]\)

⇒ \(\frac{143}{7} \times 7=143 \mathrm{~cm}\)

The total length of such a spiral made up of thirteen consecutive semicircle 143cm.

Question 19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 1 9 in the next row, 1 8 in the row next to it and so on (see figure). In how many rows are the 200 logs placed and how many logs are in the top row?

Arithmetic Progression 200 Logs Are Stacked

Solution:

No. of logs in lowest row = 20.

Starting from the bottom.

Logs in first row = 20

Logs in the second row =19

Logs in the third row = 18

This sequence is an A.P. in which

a = 20, d = 19-20 =-l

Let no. of rows = n

∴ Sn=200

⇒ \(\frac{n}{2}[2 a+(n-1) \cdot d]=200\)

⇒ \(\frac{n}{2}[2 \times 20+(n-1) \cdot(-1)]=200\)

⇒ \(n(40-n+1)=400\)

⇒ n(41-n)=400

⇒ \(41 n-n^2=400\)

⇒ \(0=n^2-41 n+400\)

⇒ \(n^2-25 n-16 n+400=0\)

⇒ \(n(n-25)-16(n-25)=0\)

⇒ \((n-25)(n-16)=0\)

⇒ \(n-25=0 \text { or } n-16=0\)

⇒ \(n=25 \text { or } n=16\)

∴ n = 25th

a25 =n + (25-1)d = 20 + 24(-l)

= -4 which is not possible.

∴ n= 16

n16 =n + (16-1)d = 20 + 1 5(-1)

= 20-15 = 5

So total rows = 1 6

and no. of logs in upper row = 5

Question 20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

Arithmetic Progression There Are Ten Potatoes In The Line

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket drops it In, and continues in the same way until all the potatoes arc in the bucket. What is the total distance the competitor has to run?
Solution:

Distance of the first potato from the first bucket = 5 m

Distance of the second potato from a bucket

= 5 + 3 = 8 m

Distance of the third potato from the bucket

= 8 + 3 = 11 cm

Once start from a bucket, pick up a potato and run back with it, drop it in the bucket.

Distance covered to drop the potatoes in the bucket.

= 2×5m,2×8m,2×11m, …

= 10 m, 16 m, 22 m,

Here a = 10, d = 16 – 10 = 22 – 16 = 6, n = 10

Distance covered to drop n potatoes in a bucket

⇒ \(\frac{n}{2}[2 a+(n-1) d]\)

Distance covered to drop 10 potatoes in a bucket

⇒ \(\frac{10}{2}[2 a+(10-1) d]\)

⇒ \(5[2 \times 10+9 \times 6]=5(20+54)\)

⇒ \(5 \times 74=370 \mathrm{~m}\)

The total distance the competitor has to run 370 meters,

Arithmetic Progression Exercise 5.4 (Optional)

Question 1. Which term of the A.P.: 121, 117, 113…..is its first negative term?

[Hint: Find n for an < 0]

Solution:

Given, A.P.: 121, 117, 113, …

Here a= 121,d= 117- 121 = 113- 117 = -4

Let \(a_n<0 \quad \Rightarrow a+(n-1) d<0\)

⇒ \(121+(n-1)(-4)<0\)

⇒ \(121-4 n+4<0\)

⇒ \(-4 n<-125\)

⇒ \(n>\frac{125}{4} \quad \Rightarrow \quad n>31 \frac{1}{4}\)

n = 32, 33, 34…

∴ The first negative term = 32nd term.

Question 2. The sum of the third and the seventh terms of an A.P. is 6 and their product is 8. Find the sum of the first sixteen terms of the A.P.
Solution:

Let the first term of A.P. be a and the common difference be d.

∴ Third term a3 = a + 2d

and Seventh term a7 = a + 6d

According to the problem,

⇒ \( a_3+a_7=6\)

⇒ \(a_3 a_7=8\)

⇒ \(a_3\left(6-a_3\right)=8 \text { [from equation }(1)]\)

⇒ \(6 a_3-a_3^2=8\)

⇒ \(0=a_3^2-6 a_3+8\)

⇒ \(a_3^2-4 a_3-2 a_3+8=0\)

⇒ \(a_3\left(a_3-4\right)-2\left(a_3-4\right)=0\)

⇒ \(\left(a_3-4\right)\left(a_3-2\right) =0\)

⇒ \(a_3-4=0 \text { or } \quad a_3-2=0\)

⇒ \(a_3=4 \text { or } \quad a_3=2\)

If a3 = 4 then from equation (1)

Now, \( a_7=6-4=2\)

a+2 d=4 ….(2)

a+6 d=2 ….(3)

On subtracting

-4d = 2

⇒ \(d=-\frac{1}{2}\)

Put the value of d in equation (2),

⇒ \(a+2\left(-\frac{1}{2}\right)=4\)

a-1=4 ⇒ a = 4+1=5

⇒ \(S_{16}=\frac{16}{2}[2 a+(16-1) d]\)

⇒ \(8\left[2 \times 5+15\left(-\frac{1}{2}\right)\right]\)

⇒ \(8\left(10-\frac{15}{2}\right)\)

⇒ \(8 \times \frac{5}{2}=20\)

If a3 = 2 then from equation (1), a7 = 4

Now, a + 2d = 2 …..(4)

a + 6d = 4 …..(5)

On subtracting

-4d =-2

d = \(\frac{1}{2}\)

Put the value of d in equation (4)

⇒ \(a+2 \times \frac{1}{2}=2 \Rightarrow a=2-1=1\)

⇒ \(S_{16}=\frac{16}{2}[2 a+(16-1) d]\)

⇒ \(8\left[2 \times 1+15 \times \frac{1}{2}\right]\)

⇒ \(8\left(2+\frac{15}{2}\right)\)

⇒ \(8 \times \frac{19}{2}=76\)

The sum of the first sixteen terms of the A.P = 76.

Question 3. A ladder has rungs 25 cm apart. (see figure). The rungs decrease uniformly in length from 45cm at the bottom to 25cm at the top. If the top and the bottom rungs are 2\(\frac{1}{2}\) m apart, what is the length of the wood required for the rungs?

[Hint : Number of rungs = \(\frac{250}{25}+1\)]

Arithmetic Progression Horizontal Distance Between First And Last Rung

Solution:

Horizontal distance between first and last rung = 2\(\frac{1}{2}\) m = 250cm

and distance between two consecutive rungs

Number of rungs in ladder \(=\frac{250}{25}+1=10+1=11\)

Now, the length of the first rung a = 25 cm

Length of last rung l = 45 cm

Length of wood used in 11 rungs

⇒ \(\frac{11}{2}(a+l)=\frac{11}{2}(25+45)\)

⇒ \(11 \times 35=385 \mathrm{~cm}\)

The length of the wood required for the rungs 385cm.

Questionfrom4. The1 tohouses49. Showofa row that is numbered there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

Hint: \(S_{x-1}=S_{49}-S_x\)

Solution:

Numbers mark on houses: 1, 2, 3, …47, 48, 49

x is a number such that Sum of the numbers before x = sum of the numbers after x

1 +2 + 3 + …+(x-1)

= (x+ 1) + (x + 2) + … + 49

⇒ \(S_{x-1}=S_{49}-S_x\)

⇒ \( \frac{x-1}{2}[1+x-1]=\frac{49}{2}[1+49]-\frac{x}{2}(1+x)\)

⇒ \(\frac{x^2-x}{2}=1225-\frac{x^2+x}{2}\)

⇒ \(\frac{x^2-x}{2}+\frac{x^2+x}{2}=1225\)

⇒ \(\frac{x^2-x+x^2+x}{2}=1225\)

⇒ \(x^2=(35)^2\)

x=35

The value of x =35.

Question 5. A small terrace at a football ground comprises of 1 5 steps each of which is 50 m long and built of solid concrete.

Each step has a rise of \(\frac{1}{4}\)m and a tread of \(\frac{1}{2}\)m (see figure). Calculate the total volume of concrete required to build the terrace.

[Hint: Volume of concrete required to build first step = \(\frac{1}{4} \times \frac{1}{2} \times 50 \mathrm{~m}^3\)]

Arithmetic Progression Volume Of Concrete Required To Build

Solution:

Given, the length of each step=50m and breadth is \(\frac{1}{2}\)m

The number of steps are 15 and the height of each step from the ground from an A.P. is as follows:

⇒ \(\frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{4}{4}, \frac{5}{4}, \frac{6}{4}, \frac{15}{4}\)

So, the volume of concrete used in the first step

⇒ \(50 \times \frac{1}{2} \times \frac{1}{4}=\frac{50}{8} \mathrm{~m}^3\)

The volume of concrete used in the second step

⇒ \(50 \times \frac{1}{2} \times \frac{2}{4}=\frac{100}{8} \mathrm{~m}^3\)

The volume of concrete used in the third step

⇒ \(50 \times \frac{1}{2} \times \frac{3}{4}=\frac{150}{8} \mathrm{~m}^3\)

The volume of concrete used in the fourth step

⇒ \(50 \times \frac{1}{2} \times \frac{4}{4}=\frac{200}{8} \mathrm{~m}^3\)

So, the volume of total concrete

⇒ \(\frac{50}{8}+\frac{100}{8}+\frac{150}{8}+\frac{200}{8}+\ldots+\text { to } 15 \text { term }\)

⇒ \(a=\frac{50}{8}\)

⇒ \(d=\frac{100}{8}-\frac{50}{8}=\frac{50}{8} \text { and } n=15\)

Therefore, the total volume of concrete

⇒ \(V=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(\frac{15}{2}\left[2 \times \frac{50}{8}+(15-1) \frac{50}{8}\right]\)

⇒ \(\frac{15}{2} \times \frac{50}{8}[2+15-1]\)

⇒ \(\frac{15}{2} \times \frac{50}{8} \times 16=15 \times 50=750 \mathrm{~m}^3\)

Therefore, the volume of concrete used in the terrace = 750 m3

Arithmetic Progression Multiple Choice Questions

Question 1. The sum of the first 6 multiples of 3 is :

  1. 55
  2. 60
  3. 63
  4. 65

Answer: 3. 63

Question 2. The sum of 10 terms of the progression 5, 11, 17, … is :

  1. 300
  2. 320
  3. 280
  4. 240

Answer: 2. 320

Question 3. 8 times the 8th term of an A.P. is equal to 12 times the 12th term. Its 20th term is

  1. 20
  2. 0
  3. -20
  4. None of these

Answer: 2. 0

Question 4. The first two terms of an A.P. are 2 and 7. Its 18th term is :

  1. 87
  2. 92
  3. 82
  4. None of these

Answer: 1. 87

Question 5. How many terms are there in A.P. 42, 63, 84 … 210?

  1. 7
  2. 8
  3. 10
  4. 9

Answer: 4. 9

Question 6. In an A.P., d =-4,n = 7,an = 7, then the value of a is:

  1. 6
  2. 7
  3. 28
  4. 30

Answer: 3. 28

Question 7. The common difference between the two arithmetic progressions are same. If their first terms are 2 and 10 respectively, then the difference between their 5th terms is :

  1. 8
  2. 2
  3. 10
  4. 6

Answer: 1. 8

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Quadratic Equation

Quadratic Equation

A quadratic equation In the variable x Is the equation of the form ax2+bx+c=0, where a, b, c are real numbers, a ≠ 0. For example.,

  1. 3x2 + 5x -1=0
  2. 3x-x2 + 1 =0

Roots Of Quadratic Equation

A real number a is called a root of the quadratic equation ax2 + bx + c = 0, a= 0 if
2 + bα + c = 0

i.e., x = α satisfies the equation ax2 + bx + c = 0

or x = α is a solution of the equation ax2 + bx + c = 0

The roots of a quadratic equation ax2 + bx + c = 0 are called zeroes of the polynomial ax2 + bx + c.

Solution Of A Quadratic Equation By Factorisation Method

Consider the quadratic equation ax2 + bx + c = 0, a≠0

Let it be expressed as a product of two linear expressions (factors) namely (px + q) and (rx + s) where p, q, r, s are real numbers and p ≠ 0, r ≠ 0, then

ax2 + bx + c = 0 ⇒ (px + q)(rx + s) = 0

⇒ px + q = 0  or  rx + s = 0

⇒ \(x=-\frac{q}{p}\)  or  \(x=-\frac{s}{r}\)

NCERT Exemplar Solutions For Class 10 Maths Chapter 4 Quadratic Equation

Solution Of A Quadratic Equation Solved Examples

Question 1. Which of the following are the solutions of 2x2 – 5x – 3 = 0?

  1. x = 2
  2. x = 3
  3. \(x=-\frac{1}{2}\)

Solution:

The given equation is 2x2 – 5x – 3 = 0

1. On substituting x = 2 in the given equation

L.H.S. = 2×22 – 5×2-3 = 8-10-3 = -5 ≠ R.H.S.

∴ x = 2 is not a solution of 2x2 – 5x – 3 = 0

2. On substituting x = 3 in the given equation

L.H.S. = 2×32– 5×3-3 = 18 -15- 3 = 0 = R.H.S. 2

∴ x = 3 is a solution of 2×2 – 5x – 3 = 0

3. On substituting x = \(-\frac{1}{2}\) in the given equation

L.H.S = \(2 \times\left(\frac{-1}{2}\right)^2-5 \times\left(\frac{-1}{2}\right)-3\)

⇒ \(2 \times \frac{1}{4}+5 \times \frac{1}{2}-3\)

⇒ \(\frac{1}{2}+\frac{5}{2}-3=\frac{1+5-6}{2}\)

=0 = R.H.S

∴ x = \(-\frac{1}{2}\) is a solution of 2×2-5x-3 = 0

Question 2. If x = 2 and x = 3 are roots of the equation 3x2-mx+ 2n = 0, then find the values of m and n.
Solution:

Since, x = 2 is a solution of 3x2 – mx + 2n = 0

∴ 3 × (2)2 -m×2 + 2n = 0

⇒ 12 -2m + 2n=0

⇒ -m + n = -6 ……(1)

Again, x = 3 is a solution of 3x2 – mx + 2n = 0

∴ 3 × (3)2 – m × 3 + 2n = 0

⇒ 27 – 3m + 2n = 0

⇒ -3m + 2n = -27 ……(2)

On multiplying equation (1) by 2 and subtracting from (2) we get

-m =-15 or m= 15

On substituting the value of m in equation (1) we get

-15 + n =-6

⇒ n= -6+15

⇒ n=9

Hence,The values of m = 15 and n = 9

Question 3. Solve the following quadratic equation : (3x-5)(2x + 3) = 0
Solution:

Given equation is (3x – 5)(2x + 3) = 0

⇒ 3x -5=0 or 2x + 3 = 0

⇒ 3x = 5 or 2x = -3

⇒ \(x=\frac{5}{3}\)  or  \(x=-\frac{3}{2}\)

Here, x = \(\frac{5}{3}\) and x = \(-\frac{3}{2}\) are the solutions.

Question 4. Find the roots of the following quadratic equation by factorization: 2z2 + az – a2 = 0
Solution:

Given equation is 2z2 + az – a2 =0

⇒ 2z2 + (2a – a) z -a2 = 0

⇒ 2z2 + 2az – az-a2 = 0

⇒ 2z(z + a) -a(z + a) = 0

⇒ (z + a)(2z – a) = 0

⇒ z + a = 0 or 2z-a = 0

when z + a = 0 ⇒ z = – a

and 2z – a = 0 ⇒ z = \(\frac{a}{2}\)

Hence, the roots of the equation are -a and \(\frac{a}{2}\)

Question 5. Solve the following quadratic equations by factorization:

  1. 4-11x = 3x2
  2. \(x^2-\frac{11}{4} x+\frac{15}{8}=0\)

Solution:

1. Given equation is 4 -11x = 3x2

⇒ 3x2+11x-4 = 0

⇒ 3x2+(12-1)-4 = 0

⇒ 3x2+12x-x-4 = 0

⇒ 3x(x+4)(x+4)

⇒ (3x-1)(x+4)=0  or x+4 = 0

when 3x+1 = 0

⇒ \(x=\frac{1}{3}\)

and x + 4 = 0

x = -4

Hence,\(\frac{1}{3}\) and- 4 are roots of equation

2. Given equation is \(x^2-\frac{11}{4} x+\frac{15}{8}=0\)

Multiplying both sides by 8, we get

⇒ 8x2– 22x + 15 = 0

⇒ 8x2 – (12 + 10)x+ 15 = 0

⇒ 8x2– 12x- 10x+ 15 = 0

⇒ 4x(2x- 3) – 5(2x- 3) = 0

⇒ (2x-3)(4x-5) = 0

∴ either 2x- 3 = 0  or 4x- 5 = 0

2x = 3 or 4x = 5

⇒ \(x=\frac{3}{2}\) or \(x=\frac{5}{4}\)

Hence \(x=\frac{3}{2}\) and \(x=\frac{5}{4}\) are the roots of given equation.

Question 6. Solve the following quadratic equation :

⇒ \(x^2-(1+\sqrt{2}) x+\sqrt{2}=0\)

Solution:

Given equation is

⇒ \(x^2-(1+\sqrt{2}) x+\sqrt{2}=0\)

⇒ \(x^2-x-\sqrt{2} x+\sqrt{2}=0\)

⇒ \(x(x-1)-\sqrt{2}(x-1)=0\)

⇒ \((x-1)(x-\sqrt{2})=0\)

x-1 = 0 or \(x-\sqrt{2}=0\)

when x-1 = 0 ⇒ x = 1

and \(x-\sqrt{2}=0\) ⇒ \(\sqrt{2}\)

Hence, 1 and \(\sqrt{2}\) are roots of the equation.

Question 7. Solve the following quadratic equation: a2b2x2 + b2x- a2x-1 = 0
Solution:

Given equation is

a2b2x2 + b2x- a2x-1=0

b2x(a2x + 1)-1 (a2x + 1 ) = 0

(a2x+1)(b2x- 1) = 0

a2x + 1 = 0 or b2x-1=0

when a2x+1=0 ⇒ \(x=-\frac{1}{a^2}\)

and b2x- 1=0 ⇒ \(x=\frac{1}{b^2}\)

Hence, \(-\frac{1}{a^2} \text { and } \frac{1}{b^2}\) are roots equation.

Question 8. Solve the following quadratic equation: 4x2– 2(a2 + b2)x + a2b2=0
Solution.

Given equation is

4x2 -2(a2 + b2)x + a2b2 = 0

4x2– 2a2x- 2b2x + a2b2 = 0

2x(2x- a2)-b2(2x -a2) = 0

(2x-2)(2x-b2) = 0

2x-a2 = 0

or 2x-b2 = 0

when 2x-a2 = 0 ⇒ \(x=\frac{a^2}{2}\)

and 2x-b2 = 0 ⇒ \(x=\frac{b^2}{2}\)

Question 9. Solve the following equation :

⇒ \(\frac{2 x}{x-4}+\frac{2 x-5}{x-3}=\frac{25}{3},(x \neq 4,3)\)

Solution:

⇒ \( \frac{2 x}{x-4}+\frac{2 x-5}{x-3}=\frac{25}{3}\)

⇒ \(\frac{2 x(x-3)+(x-4)(2 x-5)}{(x-4)(x-3)}=\frac{25}{3}\)

⇒ \(\frac{2 x^2-6 x+2 x^2-5 x-8 x+20}{x^2-3 x-4 x+12}=\frac{25}{3}\)

⇒ \(\frac{4 x^2-19 x+20}{x^2-7 x+12}=\frac{25}{3}\)

⇒ \(3\left(4 x^2-19 x+20\right)=25\left(x^2-7 x+12\right)\)

⇒ \(12 x^2-57 x+60=25 x^2-175 x+300\)

⇒ \(25 x^2-175 x+300-12 x^2+57 x-60=0\)

⇒ \(13 x^2-118 x+240=0\)

⇒ \(13 x^2-(78+40) x+240=0\)

⇒ \(13 x^2-78 x-40 x+240=0\)

⇒ 13x(x-6)-40(x-6) = 0

⇒ (x-6)-409x-6) = 0

⇒ x-6 = 0 or 13x-40 = 0

⇒ x-6 0  or \(x=\frac{40}{13}\)

Hence, 6 and \(\frac{40}{13}\) are roots of the equation.

Question 10. Solve the following equation:

⇒ \(2\left(\frac{2 x-1}{x+3}\right)-3\left(\frac{x+3}{2 x-1}\right)=5,(x \neq-3,1)\)

Solution:

Given equation is

⇒ \(2\left(\frac{2 x-1}{x+3}\right)-3\left(\frac{x+3}{2 x-1}\right)=5\)…..(1)

Let \(\frac{2 x-1}{x+3}=y\)

Hence, \(\frac{x+3}{2 x-1}=\frac{1}{y}\)

Now from equation (1)

⇒ \(2 y-\frac{3}{y}=5\)

⇒ \(2 y^2-3=5 y\)

⇒ \(2 y^2-5 y-3=0\)

⇒ \(2 y^2-(6-1) y-3=0\)

⇒ \(2 y^2-6 y+1 y-3=0\)

⇒ 2y(y-3)+1(y-3)=0

⇒ (2y+1)(y-3)=0

⇒ 2y+1=0 or y-3=0

when 2y + 1 = 0 ⇒ y = \(-\frac{1}{2}\)

and y-3 = 0 ⇒ y = 3

Substituting values of y in equation (2)

when y = \(-\frac{1}{2}\)

⇒ \(\frac{2 x-1}{x+3}=-\frac{1}{2}\)

⇒ 2(2x-1) = -1(x + 3)

⇒ 4x- 2 = -x- 3

⇒ 5x = -1

⇒ x = \(-\frac{1}{5}\)

when y = 3

⇒ \(\frac{2 x-1}{x+3}=3\)

⇒ 2x- 1 = 3(x + 3) ⇒ 2x- 1 = 3x+9

⇒ -x = 10        ⇒           x =-10

Hence, x = -10 or x = \(-\frac{1}{5}\) are roots of the equation.

Question 11. Solve the equation:

⇒\(\frac{a}{x-b}+\frac{b}{x-a}=2 \quad(x \neq b, a)\)

Solution:

Given equation is \(\frac{a}{x-b}+\frac{b}{x-a}=2\)

⇒ \(\frac{a}{x-b}+\frac{b}{x-a}=1+1\)

⇒ \(\frac{a}{x-b}-1+\frac{b}{x-a}-1=0\)

⇒ \(\frac{a-x+b}{x-b}+\frac{b-x+a}{x-a}=0\)

⇒ \((a+b-x)\left(\frac{1}{x-b}+\frac{1}{x-a}\right)=0\)

a+b -x= 0 or \(\frac{1}{x-b}+\frac{1}{x-a}=0\)

when a +b-x= 0 ⇒  x=a +b

and when  \(\frac{1}{x-b}+\frac{1}{x-a}=0\) ⇒ \(\frac{x-a+x-b}{(x-b)(x-a)}=0\)

⇒ 2x-a-b = 0

⇒ 2x= a + b

⇒ \(x=\frac{a+b}{2}\)

Alternatively,

⇒ \(\frac{1}{x-b}=-\frac{1}{x-a}\)

⇒ \(x-b=a-x\)

⇒ \(2 x=a+b\)

⇒ \(x=\frac{a+b}{2}\)

Hence, x = a + b and \(x=\frac{a+b}{2}\) arc roots of the equation.

Quadratic Formula

The roots of the quadratic equation ax2 + bx+ c= 0 where a ≠ 0 can be obtained by using the formula.

⇒ \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

Proof: Given ax2 + bx + c = 0

Dividing each term by a, we get

⇒ \(x^2+\frac{b}{a} x+\frac{c}{a}=0\)   (∵ a ≠ 0)

⇒ \(x^2+\frac{b}{a} x=-\frac{c}{a}\)  (transposing the constant)

Adding \(\left(\frac{1}{2} \text { coefficient of } x\right)^2 \text { i.e. }\left(\frac{b}{2 a}\right)^2\) on both sides, we get

⇒ \(x^2+\frac{b}{a} x+\left(\frac{b}{2 a}\right)^2=\left(\frac{b}{2 a}\right)^2-\frac{c}{a}\)

⇒ \(\left(x+\frac{b}{2 a}\right)^2=\frac{b^2}{4 a^2}-\frac{c}{a}\)

⇒ \(\left(x+\frac{b}{2 a}\right)^2=\frac{b^2-4 a c}{4 a^2}\)

Taking square root on both sides, we get

⇒ \(x+\frac{b}{2 a}= \pm \frac{\sqrt{b^2-4 a c}}{2 a}\)  (try to understand ‘±’)

⇒ \(x=\frac{-b}{2 a} \pm \frac{\sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-b+\sqrt{b^2-4 a c}}{2 a}\) or ⇒ \(x=\frac{-b-\sqrt{b^2-4 a c}}{2 a}\)

This is known as “Sridharacharya Formula” or “quadratic formula”.

Discriminant

For the quadratic equation ax2 + bx + c = 0, the expression D = b2– 4ac is called the discriminant. Roots of ax2 + bx + c = 0 are real, only when b2 – 4ac > 0, otherwise they are imaginary (not real).

Sum Of Roots And Product Of Roots

We know that the two roots of a quadratic equation ax2 + bx + c = 0 are

⇒ \(\alpha=\frac{-b+\sqrt{D}}{2 a} \quad \text { and } \quad \beta=\frac{-b-\sqrt{D}}{2 a}\) where D=b2- 4acis called the discriminant.

∴ Sum of roots :

⇒ \(\alpha+\beta=\frac{-b+\sqrt{D}}{2 a}+\frac{-b-\sqrt{D}}{2 a}=\frac{-b+\sqrt{D}-b-\sqrt{D}}{2 a}=\frac{-2 b}{2 a}=\frac{-b}{a}\)

∴ Sum of roots = \(\frac{-b}{a}=-\frac{\text { Coeff. of } x}{\text { Coeff. of } x^2}\)

Product of roots :

⇒ \(\alpha \beta=\frac{-b+\sqrt{D}}{2 a} \times \frac{-b-\sqrt{D}}{2 a}\)

⇒ \(\frac{(-b)^2-(\sqrt{D})^2}{4 a^2}=\frac{b^2-D}{4 a^2}=\frac{b^2-\left(b^2-4 a c\right)}{4 a^2}=\frac{4 a c}{4 a^2}=\frac{c}{a}\)

∴ Product of roots : \(\frac{c}{a}=\frac{\text { Constant term }}{\text { Coeff. of } x^2}\)

Sum Of Roots And Product Of Roots Solved Examples

Question 1. Find the roots of the following quadratic equation, if they exist by the method of completing the square.

⇒ \(3 x^2+4 \sqrt{3} x+4=0\)

Solution:

Given equation is

⇒ \(3 x^2+4 \sqrt{3} x+4=0\)

Dividing both sides by 3

⇒ \(x^2+\frac{4 \sqrt{3}}{3} x+\frac{4}{3}=0\)

⇒ \(x^2+\frac{4}{\sqrt{3}} x=\frac{-4}{3}\)

Adding \(\left(\frac{\text { coefficient of } x}{2}\right)^2 \text { i.e., }\left(\frac{2}{\sqrt{3}}\right)^2=\frac{4}{3}\) on both sides

⇒ \(x^2+\frac{4}{\sqrt{3}} x+\frac{4}{3}=-\frac{4}{3}+\frac{4}{3}\)

⇒ \(\left(x+\frac{2}{\sqrt{3}}\right)^2=0\)

⇒ \(\left(x+\frac{2}{\sqrt{3}}\right)=0 \quad \text { and } \quad\left(x+\frac{2}{\sqrt{3}}\right)=0\)

⇒ \(x=\frac{-2}{\sqrt{3}} \quad \text { and } \quad x=\frac{-2}{\sqrt{3}}\)

Hence, roots of the equation are \(\frac{-2}{\sqrt{3}} \text { and } \frac{-2}{\sqrt{3}}\)

Question 2. Find roots of the following quadratic equations by using the quadratic formula, if they exist.

  1. 3x2 +x-4 = 0
  2. 3x2 +x+ 4 = 0

Solution:

1. Given equation is 3x2 +x-4 = 0

On comparing with ax2 + bx+ c = 0, we get

a = 3, b = 1 and c = -4

∴ Discriminant, D = b2 – 4ac

D = (1)2 – 4 × 3 × (-4)

D = 1 +48

D = 49 > 0

Hence, the given equation has two real roots.

∴ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{-1 \pm \sqrt{49}}{6}=\frac{-1 \pm 7}{6}=\frac{6}{6}\)

⇒ \(\text { or } \frac{-8}{6}\)

⇒ \(1 \text { or } \frac{-4}{3}\)

⇒ \(x=1, \frac{-4}{3}\) are roots of the equation.

2. Given equation is 3x2 +x+ 4 = 0

On comparing with ax2 + bx + c = 0, we get

a = 3, b = 1 and c = 4

∴ Discriminant, D – b2 – 4ac

D = (1)2 – 4 × 3 × 4

D = 1 – 48

D = -47 < 0

Hence, the equation has no real roots.

Question 3. Find roots of the equation by quadratic formula : x+x -(a + 2) (a+1) = 0
Solution:

The given equation is x2 + x- (a + 2) (a +1) = 0

Comparing it with Ax2 + Bx + C = 0, we get

A = 1, B = 1 and C = -(a + 2) (a+1)

∴ \(x=\frac{-B \pm \sqrt{B^2-4 A C}}{2 A}\)

⇒ \(x=\frac{-1 \pm \sqrt{1^2-4 \times 1 \times[-(a+2)(a+1)]}}{\cdots}\)

⇒ \(x=\frac{-1 \pm \sqrt{1+4\left(a^2+3 a+2\right)}}{2}\)

⇒ \(x=\frac{-1 \pm \sqrt{1+4 a^2+12 a+8}}{2}\)

⇒ \(x=\frac{-1 \pm \sqrt{4 a^2+12 a+9}}{2}\)

⇒ \(x=\frac{-1 \pm \sqrt{(2 a+3)^2}}{2}\)

⇒ \(x=\frac{-1 \pm(2 a+3)}{2}\)

⇒ \(x=\frac{-1+2 a+3}{2} \text { and } \frac{-1-2 a-3}{2}\)

⇒ \(x=\frac{2 a+2}{2} \text { and } \frac{-2 a-4}{2}\)

⇒ x = (a + 1) and -(a + 2) are roots of the equation.

Question 4. Solve the following equation by the method of completing the square :

⇒ \(4 \sqrt{3} x^2+5 x-2 \sqrt{3}=0\)

Solution:

We have, \(4 \sqrt{3} x^2+5 x-2 \sqrt{3}=0\)

Dividing each term by \(4 \sqrt{3}\) we get

⇒ \(x^2+\frac{5}{4 \sqrt{3}} x-\frac{2 \sqrt{3}}{4 \sqrt{3}}=0\)

⇒ \(x^2+\frac{5}{4 \sqrt{3}} x=\frac{1}{2}\)

Adding \(\left(\frac{1}{2} \text { coefficient of } x\right)^2 \text { i.e., }\left(\frac{5}{8 \sqrt{3}}\right)^2\) to both sides, we get

⇒ \(x^2+\frac{5}{4 \sqrt{3}} x+\left(\frac{5}{8 \sqrt{3}}\right)^2=\left(\frac{5}{8 \sqrt{3}}\right)^2+\frac{1}{2}\)

⇒ \(\left(x+\frac{5}{8 \sqrt{3}}\right)^2=\frac{25}{192}+\frac{1}{2} \Rightarrow\left(x+\frac{5}{8 \sqrt{3}}\right)^2=\frac{25+96}{192}\)

⇒ \(\left(x+\frac{5}{8 \sqrt{3}}\right)^2=\frac{121}{192}\)

Taking the square root of both sides, we get

⇒ \(x+\frac{5}{8 \sqrt{3}}= \pm \frac{11}{8 \sqrt{3}}\)

∴ \(x=-\frac{5}{8 \sqrt{3}} \pm \frac{11}{8 \sqrt{3}}=\frac{-5 \pm 11}{8 \sqrt{3}}\)

⇒ \(x=\frac{-5+11}{8 \sqrt{3}} \quad \text { or } \quad x=\frac{-5-11}{8 \sqrt{3}}\)

⇒ \(x=\frac{3}{4 \sqrt{3}} \quad \text { or } \quad x=\frac{-2}{\sqrt{3}}\)

Hence x = \(\frac{3}{4 \sqrt{3}} \text { or } x=\frac{-2}{\sqrt{3}}\) are the solutions of given equation.

Question 5. Solve: x2 + x- (a + 2) (a + 1) = 0 by

  1. Factorisation
  2. Method of completing the square

Solution:

1. By factorisation :

We have x2 +x- (a + 2) (a + 1) = 0

⇒ x2+x× 1 -{a + 2) (a+ 1) = 0

⇒ x2 +x[(a + 2) — (a + 1)] — (a + 2) (n + 1) = 0

⇒ x2 +x(a + 2) -x(a + 1) – (a + 2) (a + 1) = 0

⇒ x[x+ (a + 2)] – (a + 1)[x+ (a + 2)] = 0

⇒ [x+(a + 2)] [x-(a+ 1)] = 0

∴ either x+ (a + 2) = 0  or x- (a + 1) = 0

⇒ x = -(a + 2) or x=(a+ 1)

2. By the method of completing the square:

We have x2 + x- (a + 2) (a + 1) = 0

⇒ x2 +x=(a + 2) (a + 1)

Adding \(\left(\frac{1}{2}\right)^2\) on both sides, we get

⇒ \(x^2+x+\left(\frac{1}{2}\right)^2=a^2+3 a+2+\left(\frac{1}{2}\right)^2\)

⇒ \(\left(x+\frac{1}{2}\right)^2=\frac{4 a^2+12 a+9}{4}\)

⇒ \(\left(x+\frac{1}{2}\right)^2=\left(\frac{2 a+3}{2}\right)^2\)

Taking the square root of both sides, we get

⇒ \(x+\frac{1}{2}= \pm \frac{2 a+3}{2}\)

∴ \(x=\frac{-1}{2} \pm \frac{2 a+3}{2}=\frac{-1 \pm(2 a+3)}{2}\)

∴ \(x=\frac{-1+2 a+3}{2}\text { or }x=\frac{-1-(2 a+3)}{2}\)

⇒ \(x=\frac{2(a+1)}{2}\text { or }x=\frac{-2(a+2)}{2}\)

⇒ \(x=(a+1)\text { or }x=-(a+2)\)

Question 6. Let f(x) = 3x2 – 5x- 1. Then solve f(x) = 0 by

  1. Factoring the quadratic
  2. Using the quadratic formula
  3. Completing the square and then rewrite f(x) in the formv4(x±B)2 ± C.

Solution:

f(x) = 3x2– 5x- 1

f(x) = 0

3x2-5x- 1=0

1. The given quadratic equation cannot be fully factorised using real integers. So, it is better to solve this equation by any other method.

2. 3x-5x- 1 =0

Compare it with ax2 +bx + c = 0, we get

a = 3, b = -5,c =-1

∴ Let two roots of this equation are

⇒ \(\alpha=\frac{-b+\sqrt{b^2-4 a c}}{2 a}\)

⇒ \(\frac{-(-5)+\sqrt{(-5)^2-4(3)(-1)}}{2 \times 3}\)

⇒ \(\frac{5+\sqrt{37}}{6}\)

and

⇒ \(\beta=\frac{-b-\sqrt{b^2-4 a c}}{2 a}\)

⇒ \(\frac{-(-5)-\sqrt{(-5)^2-4(3)(-1)}}{2 \times 3}\)

⇒ \(\beta=\frac{5-\sqrt{37}}{6}\)

∴ Two values of x are \(\frac{5+\sqrt{37}}{6} { and } \frac{5-\sqrt{37}}{6}\)

3. \(3 x^2-5 x-1=0\)

⇒ \(x^2-\frac{5}{3} x-\frac{1}{3}=0\)

⇒ \(x^2-\frac{5}{3} x+\ldots \ldots=\frac{1}{3}+\ldots \ldots\)

⇒ \(x^2-\frac{5}{3} x+\left(\frac{5}{6}\right)^2\)

⇒ \(\frac{1}{3}+\left(\frac{5}{6}\right)^2\)

⇒ \(\left[\text { adding }\left(\frac{\text { Coeff. of } x}{2}\right)^2 \text { on both sides }\right]\)

⇒ \(\left(x-\frac{5}{6}\right)^2=\frac{1}{3}+\frac{25}{36} \Rightarrow\left(x-\frac{5}{6}\right)^2=\frac{12+25}{36}\)

∴ \(\left(x-\frac{5}{6}\right)^2=\left(\frac{\sqrt{37}}{6}\right)^2\)

⇒ \(x-\frac{5}{6}= \pm \frac{\sqrt{37}}{6}\)

∴ Two values of x are \(\frac{5+\sqrt{37}}{6} \text { and } \frac{5-\sqrt{37}}{6} \text {. }\)

Now, f(x) = \(3 x^2-5 x-1=3\left(x^2-\frac{5}{3} x\right)-1\)

⇒ \(3\left(x^2-\frac{5}{3} x+\frac{25}{36}-\frac{25}{36}\right)-1\)

⇒ \(3\left(x-\frac{5}{6}\right)^2-\frac{25}{12}-1=3\left(x-\frac{5}{6}\right)^2-\frac{37}{12}\)

which is of the form A(x- B)2– C, where A = 3, B = \(\frac{5}{6}\), C = \(\frac{37}{12}\)

Nature Of Roots Of A Quadratic Equation

The nature of roots of a quadratic equation ax2 +bx + c = 0 depends on the value of its discriminant (D). i.e., upon b2 – 4ac.

If a, b, and c are real numbers and a ≠ 0 then discriminant D = b2-4ac.

The value of discriminant affects the nature of roots in the following ways :

The roots of a quadratic equation are :

  1. Real: When D > 0 i.e. (D > 0 or D = 0) (when quadratic equation can be expressed as in real linear factors, D > 0)
  2. No Real (Imaginary): When D < 0
  3. Real and Distinct: When D > 0
  4. Real and Equal (Coincident): When D = 0

In this case each equal root will \(\left(\frac{-b}{2 a}\right)\)

Remember:

  1.  ax-b>0 ⇒ \(x>\frac{b}{a} \text {, if } a>0 \text { and } x<\frac{b}{a} \text {, if } a<0\)
  2. x2– a2 > 0 ⇒ x<-a or x>a
  3. x2– a2 = 0 ⇒ x= —a or x =  a
  4. x2– a2 < 0 ⇒ x<a or x> -a ⇒ -a< x < 0
  5. (x-a) (x-b)>0,a<b ⇒ x<a or x>b
  6. (x- a) (x- b) < 0, a < b ⇒ a<x<b

Solved Examples

Question 1. Find the value of k so that the equation 2x2 – 5x + k = 0 has two equal roots.
Solution:

The given equation is 2x2 – 5x + K = 0

Comparing with ax2 + bx + c = 0, we get

a = 2,b = -5 and c = k

The equation will have two equal roots if

D = 0

D = b2– 4ac = 0

or (-5)2 -4×2×K = 0

25 – 8K = 0

⇒ \(k=\frac{25}{8}\)

The value of k =\(\frac{25}{8}\)

Question 2. The equation 3x2 – 12x + (n – 5) = 0 has repeated roots. Find the value of n.
Solution:

Given equation is 3x2– 12x + (n – 5) = 0

Comparing with ax2 +bx + c = 0, we get

a = 3, b = -12 and c = n- 5

The equation will have repeated (two equal) roots if

Discriminant (D) = 0

∴ D =b2-4ac=0

or (-12)2– 4 × 3 × (n- 5) = 0

⇒ 144 -12n + 60 = 0

⇒ 204- 12n = 0

⇒ -12n = -204

⇒ n = 17

Hence, the value of n is 17

Question 3. Find the value of k for which the equation 2 + k(2x +k- l) + 2 = 0 has real and equal
roots.
Solution:

Given equation is

x2 + k(2x + k – 1) + 2 = 0

x2 + 2kx +k(k- 1) + 2 = 0

x2 + 2kx + (k2 -k + 2) = 0

Comparing with ax2 + bx + c = 0,we get

a = 1, b = 2k and c = k2 – k + 2

For real and equal roots,

Discriminant (D) = 0

∴ D =b2-4ac=0

or (2k)2 – 4(1)(k2 -k + 2)= 0

4k2 – 4(k2 -k + 2) = 0

k2– (k2 -k + 2) = 0

K-2 = 0

k = 2.

Hence, the value of k is 2

Question 4. Find the value of p, for which one root of the quadratic equation px2 – 14x + 8 = 0 is 6 times the other.
Solution:

Given equation is px2 – 14x + 8 = 0

Let a and p be two roots of quadratic equation px2– 14x + 8 = 0, such that β = 6a

∴ Sum of roots = \(\frac{-b}{a}\)

⇒ \(\alpha+6 \alpha=-\frac{(-14)}{p}\)

⇒ \(7 \alpha=\frac{14}{p} \Rightarrow \alpha=\frac{2}{p}\)….(1)

Product of roots = \(\frac{c}{a}\)

⇒ \(\alpha \cdot 6 \alpha=\frac{8}{p}\)

⇒ \(6 \alpha^2=\frac{8}{p} \Rightarrow \alpha^2=\frac{8}{6 p}\) ….(2)

From equations (1 ) and (2), we get

⇒ \(\left(\frac{2}{p}\right)^2=\frac{8}{6 p} \Rightarrow \frac{4}{p^2}=\frac{4}{3 p}\)

P2 = 3P

p2 – 3p = 0

p(p-3) = 0 (don’t cancel p on both sides)

Either p = 0 or p = 3.

But p = 0 is not possible, as on putting, p = 0 in the given equation, we don’t have a quadratic equation and therefore we cannot get two roots.

Hence, P = 3

Alternatively,

Let one root of the quadratic equation px2 – 14x + 8 = 0 is a.

∴ pα2 – 14α + 8 =0 ….(1)

∴ Other root of the equation will be 6a.

p(6α)2– 14(6α) + 8 =0

36pα2– 84α + 8=0

9pα2 – 21α + 2 =0 ….(2)

Solving equations (1) and (2) by cross-multiplication method.

⇒ \(\frac{\alpha^2}{-14(2)-8(-21)}=\frac{\alpha}{8(9 p)-2 p}=\frac{1}{p(-21)-9 p(-14)}\)

⇒ \(\frac{\alpha^2}{-28+168}=\frac{\alpha}{70 p}=\frac{1}{105 p}\)

⇒ \( \frac{\alpha^2}{140}=\frac{1}{105 p}\text { and }\frac{\alpha}{70 p}=\frac{1}{105 p}\)

⇒ \(\alpha^2=\frac{140}{105 p}=\frac{4}{3 p}\text { and }\alpha=\frac{70 p}{105 p}=\frac{2}{3}\)

⇒ \(\left(\frac{2}{3}\right)^2=\frac{4}{3 p} \quad \Rightarrow \quad \frac{4}{9}=\frac{4}{3 p} \quad \Rightarrow \quad 3 p=9\)

p = 3

The value of p = 3

Question 5. The equation x2+2(m-1)x+ (m + 5) 0 has real and equal roots. Find the value of m.
Solution:

Given equation is x2+ 2 (m – 1 )x + (m+ 5) = 0

Comparing with ax2 + bx + c = 0, we get

a = 1, b = 2(m -1), c = m + 5

The equation will have two real and equal roots if

Discriminant (D) = 0

∴ D = b2 – 4ac= 0

or [2(m-1)]2-4 × 1 × (m + 5) = 0

4(m2 + 1 – 2m) – 4(m + 5) = 0

4m2 + 4- 8m – 4m – 20 = 0

4m2– 12m, – 16 = 0

m2– 3m, -4 = 0

m2 – 4m, + m, – 4 = 0

m(m, – 4) + 1 (m – 4) = 0

(m + 1) (m – 4) = 0

m = -1 or m = 4

Hence, the value(s) of m are -1 and 4.

Question 6. If -4 is a root of the equation x2 + px- 4 = 0 and the equation x2 + px + q = 0 has coincident roots, find the values of p and q.
Solution:

Since -4 is a root of x2 + px- 4 = 0

Hence, (-4) will satisfy the equation.

Therefore, (-4)2 +p(-4) -4=0

16 – 4p – 4 = 0

-4p +12 = 0

-4p = -12 .

p = 3 …….(1)

Given that, x2 + px + q = 0 has coincident roots.

D = b2 – 4ac = 0

D = p2-4×1×q=0

p2-4q = 0

32 -4q =0 [from (1)]

9 – 4q = 0

-4q = -9

q = \(\frac{9}{4}\)

Hence, the values of p = 3 and q = \(\frac{9}{4}\)

Question 7. Prove that both roots of the equation (x -a) (x- b) + (x- b) (x- c) + {x- c) (x-a) = 0 are real but they are equal only when a =b =c.
Solution:

The given equation may be written as

3x2 – 2(a + b + c)x+ (ab + bc + ac) = 0

∴ Discriminant D = B2 – 4AC

D = \(4(a+b+c)^2-4 \times 3(a b+b c+a c)\)

D = \(4\left(a^2+b^2+c^2+2 a b+2 b c+2 a c\right)-12(a b+b c+a c)\)

D = \(4\left(a^2+b^2+c^2-a b-b c-a c\right)\)

D = \(2\left(2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 a c\right)\)

D = \(2\left[a^2+b^2-2 t b+b^2+c^2-2 b c+c^2+a^2-2 a c\right]\)

D = \(2\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \geq 0\)

∵ \((a-b)^2 \geq 0,(b-c)^2 \geq 0 \text { and }(c-a)^2 \geq 0\)

Hence, both roots of the equation are real.

For equal root, we must have D = 0

⇒ (a – b)2 + (b- c)2 + (c- a)2 = 0

⇒ a-b =0, b-c = 0, c-a = 0

⇒ a = b,b = c,c = a

⇒ a – b =c

Hence, roots are equal only when a=b = c

Question 8. Find the positive values of k for which the equations x2 + kx + 64 = 0 and x2 – 8x + k = 0 nail both have real roots:
Solution:

Given equations are

x2 + kx + 64 = 0 …(1)

and x2-8x + k= 0 …(2)

Let D1 and D2 be discriminants of equations (1) and (2) respectively, then

D1 = k2 – 4 x 64 or  D1 = k2 – 656

and D2 = (-8)2 – 4k or D2 = 64-4K

Both equations will have real roots, if

D1 ≥ 0 and D2 ≥ 0

⇒ K2 -256 ≥ 0 and 64- 4k ≥ 0

⇒ k2 ≥ 256 and64 ≥ 4k

⇒ k ≥ 16 and K ≤ 16

k = 16

Hence, both equations will have real roots, when k = 16.

The positive values of k = 16.

Question 9. Find the value(s) of k for which the given quadratic equations have real and distinct roots:

  1. 2x2 + kx + 4 = 0
  2. 4x2-3kx+ 1=0
  3. kx2 + 6x + 1 = 0
  4. x2-kx+ 9 = 0

Solution:

1. The given equation is 2x2 + kx + 4 = 0

Comparing with ax2 +bx + c = 0, we get

a-2, b=k and c = 4

∴ D = b2– 4ac ≥ 0 for real and distinct roots.

Therefore, D = k2– 4 × 2 × (4) ≥ 0

⇒ k2 – 32 ≥ 0

⇒ k2 ≥ 32

⇒ \(k \leq-4 \sqrt{2} \text { and } k \geq 4 \sqrt{2}\)

2. The given equation is 4x2 – 3kx + 1 – 0

Comparing with ax2 + bx + c = 0, wc get

a = 4, b = -3k and c = 1

∴ D = b2 – 4ac ≥ 0 for real and distinct roots

Therefore, D = (-3k)2 – 4 × 4 × 1 ≥ 0

9k2 -16>0 ⇒ 9k2 > 16

⇒ \(k^2 \geq \frac{16}{9}\)

⇒ \(k \leq-\frac{4}{3} \quad \text { and } \quad k \geq \frac{4}{3}\)

3.  The given equation is kx2 + 6x+1 = 0

Comparing with ax2 +bx + c = 0, we get

a = k, b = 6 and c = 1

D = b2 – 4ac > 0 for real and distinct roots

Therefore, D = (6)2– 4 × k × 1 ≥ 0

36-4K ≥ 0

36 ≥ 4K

k ≤ 9

4. The given equation is

x2 -kx + 9 = 0

Comparing with ax2 + bx + c = 0, we get

a = 1, b = -k and c = 9

∴ D = b2 – 4ac ≥ 0 for real and distinct roots

Therefore, D = (-k)2– 4 × 1 × 9 ≥ 0

⇒ K2 – 36 ≥ 0

⇒ k ≤ -6 or K ≥ 6

Question 10. If roots of the equation (1 + m2)x2 + 2mcx + (c2 – a2) = 0 are equal, prove that : c2 = a2(1+m2)
Solution:

We have,

(1 + m2)x2 + 2mcx + (c2 – a2) = 0

It has equal roots, if

D = 0

B2-4AC =0

⇒ (2mc)2 – 4(1 + m2) (c2 – a2) =0

⇒ 4m2c2 – 4(c2 – a2 + m2c2– m2a2) = 0

⇒ m2c2 -c2 + a2 – m2c2 + m2a2 = 0

⇒ c2 = a2+ m2a2 = a2(1 +m2)

Word Problems Based On Quadratic Equations

To solve the word problem, first translate the words of the problem into an algebraic equation, then solve the resulting equation.

For solving a word problem based on a quadratic equation adopt the following steps:

Step 1: Read the statement of the problem carefully.

Step 2: Represent the unknown quantity of the problem by a variable.

Step 3: Translate the given statement to form an equation in terms of variables.

Step 4: Solve the equation.

Quadratic Equations Solved Examples

Question 1. The sum of a number and its reciprocal is \(\frac{10}{3}\), find the number(s).
Solution:

Let the number be x

∴ According to a given statement

⇒ \(x+\frac{1}{x}=\frac{10}{3}\)

⇒ \(\frac{x^2+1}{x}=\frac{10}{3}\)

⇒ \(3 x^2+3=10 x\)

⇒ \(3 x^2-10 x+3=0\)

⇒ \(3 x^2-(9+1) x+3=0\)

⇒ \(3 x^2-9 x-x+3=0\)

⇒ 3x(x-3)- l(x-3) = 0

⇒ (3x-1) (x-3) = 0

when 3x- 1 = 0  \(x=\frac{1}{3}\)

and when x-3 = 0,  x = 3

Hence, the number(s) are 3 and \(\frac{1}{3}\)

Question 2. Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.
Solution:

Let the larger part be x. Then, the smaller part = 16 – x

According to a given statement

⇒ 2x2-(16-x)2 = 164

⇒ 2x2 – (256 +x2 – 32x) = 164

⇒ 2x2 – 256- x2 + 32x- 164 = 0

⇒ x2 + 32x- 420 = 0

⇒ x2 + 42x- 10x- 420 = 0

⇒ x(x + 42)- 10(x + 42) = 0

⇒ (x + 42) (x- 10) = 0

⇒ x = -42 or x =10

⇒ x = 10

Hence, the required parts are 10 and 6.

Question 3. The sum of squares of three consecutive natural numbers is 149, Find the numbers.
Solution:

Given

The sum of squares of three consecutive natural numbers is 149,

Let three consecutive natural numbers be a-, (x +1) and (x +2) respectively.

According to the given condition

⇒ \(x^2+(x+1)^2+(x+2)^2=149\)

⇒ \(x^2+\left(x^2+1+2 x\right)+\left(x^2+4+4 x\right)=149\)

⇒ \(3 x^2+6 x+5=149\)

⇒ \(3 x^2+6 x-144=0\)

⇒ \(x^2+2 x-48 =0\)

⇒ \(x^2+8 x-6 x-48=0\)

⇒ x(x+8)-6(x+8)=0

⇒ (x+8)(x-6)=0

⇒ x = -8 and x = 6

⇒ x = 6 (x = -8 is not a natural number)

Hence, the required natural numbers are 6, (6 + 1), (6 + 2) = 6, 7, 8 respectively.

Question 4. A two-digit number is such that the product of its digits is 8. When 63 is subtracted from the number digits interchange their places. Find the number.
Solution:

Given

A two-digit number is such that the product of its digits is 8. When 63 is subtracted from the number digits interchange their places.

Let the digit at the unit place be x and the digit at ten’s place be y

∴ Number = x × 1 + 10 × y

= x+ 10y.

After reversing the order of digits, the reversing number =y + 10x

According to the first condition,

xy = 8 ….(1)

According to the second condition,

⇒ (x+10y) – 63 -y + 10ar

⇒ 9y- 9x = 63

⇒ y -x = 7

⇒ y =x + 7 ….(2)

∴ from (1) and (2), we get

x (x + 7) = 8

⇒ x2 +7x- 8 = 0

⇒ (x + 8) (x- 1) = 0

∴ x= 1 or x = -8

If \(\left.\begin{array}{l}
x=1 \\
y=1+7=8
\end{array}\right\} \quad \Rightarrow \text { number }=1+10(8)=81\)

If \(\left.\begin{array}{rl}
x=-8 \\
y=-8+7=-1
\end{array}\right\}\) not possible as digits cannot be negative.

So, required number = 81

Question 5. The denominator of a fraction is one more than twice the numerator. If the sum of the 16 fractions and its reciprocal is \(2 \frac{16}{21}\), find the fraction.
Solution:

Given

The denominator of a fraction is one more than twice the numerator. If the sum of the 16 fractions and its reciprocal is \(2 \frac{16}{21}\)

Let, the fraction be \(\frac{x}{y}\) where numerator is.v, then denominator = 2x + y

According to a given statement

⇒ \(\frac{x}{2 x+1}+\frac{2 x+1}{x}=2 \frac{16}{21}\)

Now, let \(\frac{x}{2 x+1}=a\)

⇒ \(a+\frac{1}{a}=\frac{58}{21} \Rightarrow \frac{a^2+1}{a}=\frac{58}{21}\)

⇒ \(21 a^2+21=58 a\)

⇒ \(21 a^2-58 a+21=0\)

⇒ \(21 a^2-49 a-9 a+21=0\)

⇒ \(7 a(3 a-7)-3(3 a-7)=0\)

⇒ \((7 a-3)(3 a-7)=0\)

⇒ \(a=\frac{3}{7} \quad \text { or } \quad a=\frac{7}{3}\)

when, \(a=\frac{3}{7} \Rightarrow \frac{x}{2 x+1}=\frac{3}{7}\)

⇒ 7x = 6x + 3

⇒ x = 3

when, \(a=\frac{7}{3} \Rightarrow \frac{x}{2 x+1}=\frac{7}{3}\)

⇒ 3x = 14x + 7

⇒ -11x = 7

⇒ \(\frac{-7}{11}\)

x = 3   \(\left(x=\frac{-7}{11} \text { is not a natural number }\right)\)

Hence, required fraction is \(\frac{x}{2 x+1}=\frac{3}{7} \text {.}\)

Question 6. The hypotenuse of a right-angled triangle is 6 meters more than twice the shorter side. If the third side is 2 meters less than the hypotenuse, find the sides of the triangle.
Solution:

Given

The hypotenuse of a right-angled triangle is 6 meters more than twice the shorter side. If the third side is 2 meters less than the hypotenuse

Let, the length of the shortest side be x meters

then hypotenuse = (2x + 6) metres

the third side = (2x + 6 – 2) metres

= (2x + 4) metres

Now, using Pythagoras theorem (2x+6)2=x2 + (2x + 4)2

⇒ (4x2 + 24x + 36)=x2 + (4x2+ 16x + 16)

⇒ x2 – 8x- 20 = 0

⇒ x2 – 10x+2x- 20 = 0

⇒ x(x-10)+2(x+2) = 0

⇒ (x+10)(x+2) = 0

⇒ x = 10 or x = -2

⇒ x = 10

So, the length of the shortest side = 10 meters

length of the hypotenuse = (2 × 10 + 6) = 26 metres

and length of the third side = (2 × 10 + 4) = 24 metres

Hence, the sides of the triangle are 10 m, 24 m, and 26 m.

Question 7. The age of a father is twice the square of the age of his son. Eight years hence, the age of his father will be 4 years more than 3 times the age of the son. Find their present ages.
Solution:

Given

The age of a father is twice the square of the age of his son. Eight years hence, the age of his father will be 4 years more than 3 times the age of the son.

Let, the present age of the son be x years.

Hence, the age of father = 2×2

8 years hence, the age of son = (x + 8) years

and the age of father = (2x2 + 8) years

According to a given statement

⇒ \(2 x^2+8=3(x+8)+4\)

⇒ \(2 x^2+8=3 x+24+4\)

⇒ \(2 x^2-3 x-20=0\)

⇒ \(2 x^2-8 x+5 x-20=0\)

⇒ \(2 x(x-4)+5(x-4)=0\)

⇒ (2 x+5)(x-4)=0

⇒ \(x=\frac{-5}{2}\) and x = 4

⇒ x = 4

Therefore, the present age of the son is 4 years and present age of the father is 2 × 42 = 32 years.

Question 8. Two taps running together can fill a tank in 3 \(3 \frac{1}{13}\) hours. If one tap takes 3 hours more 13 than the other to fill the tank, then how much time will each tap take to fill the tank?
Solution:

Given

Two taps running together can fill a tank in 3 \(3 \frac{1}{13}\) hours. If one tap takes 3 hours more 13 than the other to fill the tank

Quadratic Equations Two Taps Running Together

Let the time taken by TapI to fill a tank = x hrs

∴ The time taken by TapII to fill a tank = (x + 3) hrs

and time taken by both to fill a tank =  \(3 \frac{1}{13}=\frac{40}{13} \mathrm{hrs}\)

∴ Tap 1’s 1 hr work = \(\frac{1}{x}\)

Tap 2’s 1 hr work = \(\frac{1}{x+3}\)

and (Tap1 + Tap 2)’s 1hr work = \(\frac{13}{40}\)

⇒ \(\frac{1}{x}+\frac{1}{x+3}=\frac{13}{40}\)

⇒ \(\frac{x+3+x}{x(x+3)}=\frac{13}{40}\)

⇒ \(40(2 x+3)=13 x(x+3)\)

⇒ \(80 x+120=13 x^2+39 x\)

⇒ \(13 x^2-41 x-120=0\)

⇒ \(13 x^2-65 x+24 x-120=0\)

⇒ \(13 x(x-5)+24(x-5)=0\)

⇒ (x-5)(13 x+24)=0

∴ Either x-5 = 0 or 13x+ 24 = 0

x = 5 or \(x=\frac{-24}{13}\)

But time cannot be negative, so we reject x = \(\frac{-24}{13}\)

x = 5

Hence, time taken by Tap 1 = 5 hrs

and time taken by Tap 2 = (5 + 3) hrs = 8 hrs.

Question 9. A takes 6 hours less than B to complete a work. If together they complete the work in 13 hours 20 minutes, find how much time B alone takes to complete the work.
Solution:

Given

A takes 6 hours less than B to complete a work. If together they complete the work in 13 hours 20 minutes,

Let, B alone complete the work in* hours, then A alone will complete the work in (x- 6) hours.

⇒ \(\frac{1}{x-6}+\frac{1}{x}=\frac{3}{40}\)    \(\left(13 \mathrm{hr}+20 \mathrm{~min}=\frac{40}{3} \mathrm{hrs} .\right)\)

⇒ \(\left(13 \mathrm{hr}+20 \mathrm{~min}=\frac{40}{3} \mathrm{hrs} .\right)\)

⇒ \(\frac{x+x-6}{(x-6) x}=\frac{3}{40}\)

⇒ \(3 x^2-18 x=80 x-240\)

⇒ \(3 x^2-98 x+240=0\)

⇒ \(3 x^2-90 x-8 x+240=0\)

⇒ \(3 x(x-30)-8(x-30)=0\)

⇒ \((x-30)(3 x-8)=0\)

x = 30

⇒ \(x=\frac{8}{3}\)

x = 30   \(\left(\text { if } x=\frac{8}{3} \text {, then } x-6 \text { in negative }\right)\)

B alone will take 30 hours to complete the work.

Question 10. An airplane takes 1 hour less for a journey of 1200 km if its speed is increased by 100 km/hr from its usual speed. Find its usual speed.
Solution:

Given

An airplane takes 1 hour less for a journey of 1200 km if its speed is increased by 100 km/hr from its usual speed.

Let the usual speed of an airplane be x km/hr. Given, distance = 1200 km

Time taken for journey of 1200 km = \(\frac{1200}{x} \text { hours }\)

⇒ \(\left(\text { Time }=\frac{\text { Distance }}{\text { Speed }}\right)\)

When the speed is increased by 100 km/hr, time taken for the same journey = \(\frac{1200}{x+100} \text { hours. }\)

According to the given condition

⇒ \(\frac{1200}{x}-\frac{1200}{x+100}=1\)

⇒ \(\frac{1200(x+100)-1200 x}{x(x+100)}=1\)

⇒ \(1200 x+120000-1200 x=x^2+100 x\)

⇒ \(x^2+100 x-120000=0\)

⇒ \(x^2+400 x-300 x-120000=0\)

⇒ \(x(x+400)-300(x+400)=0\)

⇒ \((x+400)(x-300)=0\)

⇒ x = -400 or x = 300

⇒ x = 300 km/hr

∴ The usual speed = 300 km/hr.

Question 11. A motor boat whose speed is 15 km/hr in still water goes 30 Ion downstream and comes back in a total of 4 hours 30 minutes. Determine the speed of water.
Solution:

Given

A motor boat whose speed is 15 km/hr in still water goes 30 Ion downstream and comes back in a total of 4 hours 30 minutes.

Let, the speed of water be x km/hr.

Given the speed of a motor boat in still water is 15 km/hr.

Therefore, its speed downstream is (15 + x) km/hr and the speed upstream is (15- x) lon/hr.

Time taken for going 30 Ion downstream = \(\frac{30}{15+x} \text { hours. }\)

Time taken for going 30 Ion upstream = \(\frac{30}{15-x} \text { hours. }\)

According to the given condition

⇒ \(\frac{30}{15+x}+\frac{30}{15-x}=4+\frac{30}{60}\)

⇒ \(\frac{30(15-x)+30(15+x)}{(15+x)(15-x)}=\frac{9}{2}\)

⇒ \(\frac{450-30 x+450+30 x}{225-x^2}=\frac{9}{2}\)

⇒ \(900 \times 2=9\left(225-x^2\right)\)

⇒ \(1800=2025-9 x^2\)

⇒ \(9 x^2=225 \text { or } x^2=25\)

⇒ \(x= \pm 5\)

⇒ x=5

Hence, the speed of water is 5 Km/hr.

Question 12. A dealer sells an article for ₹24 and gains as much percent as the price of the article. Find the cost price of the article.
Solution:

Given

A dealer sells an article for ₹24 and gains as much percent as the price of the article.

Let, the C.P. of article be ₹x

Then, gain = x %

∴ \(\text { S.P. }=\frac{100+\text { gain } \%}{100} \times \text { C.P. } \quad\left(\text { or S.P. }=\text { C.P. }+ \text { C.P. } \times \frac{x}{100}\right)\)

⇒ \(24=\frac{100+x}{100} \times x\)

⇒ \(2400=100 x+x^2\)

⇒ \(x^2+100 x-2400=0\)

⇒ \(x^2+120 x-20 x-2400=0\)

⇒ \(x(x+120)-20(x+120)=0\)

⇒ \((x-20)(x+120)=0\)

⇒ x = 20 or x = -120

⇒ x = 20

Hence, the cost of the article is ₹20.

Question 13. A piece of cloth costs ₹200. If the piece was 5 m longer and each meter of cloth cost 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is the original rate per meter?
Solution:

Given

A piece of cloth costs ₹200. If the piece was 5 m longer and each meter of cloth cost 2 less, the cost of the piece would have remained unchanged.

Let, the length of the piece be A meters

Since the cost of A meters of cloth = ₹200

⇒ Cost of each metre of cloth = \(₹ \frac{200}{x}\)

New length of cloth = (x + 5)m

New cost of each metre of cloth = \(₹ \frac{200}{x+5}\)

Now, given \(\frac{200}{x}-\frac{200}{x+5}=2\)

⇒ \(\frac{200(x+5)-200 x}{x(x+5)}=2 \quad \Rightarrow \quad \frac{200 x+1000-200 x}{x^2+5 x}=2\)

⇒ \(1000=2 x^2+10 x\)

⇒ \(2 x^2+10 x-1000=0\)

⇒ \(x^2+5 x-500=0\)

⇒ \(x^2+25 x-20 x-500=0\)

⇒ \(x(x+25)-20(x+25)=0\)

⇒ \((x-20)(x+25)=0\)

x = 20 or x = -25

Since length cannot be negative

Hence, x = 20 m

and the original rate per metre = \(₹ \frac{200}{20}=₹ 10\)

Question 14. Some students planned a picnic. The budget for the food was ₹480. As eight of them failed to join the picnic, the cost of the food for each member increased by ₹10. Find how many students went for a picnic.
Solution:

Given

Some students planned a picnic. The budget for the food was ₹480. As eight of them failed to join the picnic, the cost of the food for each member increased by ₹10.

Let, no. of students who planned the picnic = x

Given, the budget for food = ₹480

∴ Share of each student = \(₹ \frac{480}{x}\)

Since eight of them failed to join the picnic

∴ No. of students went for picnic = (x- 8)

∴ Share of each student = \(₹ \frac{480}{x-8}\)

According to the given condition

⇒ \(\frac{480}{x-8}-\frac{480}{x}=10\)

⇒ \(\frac{480 x-480(x-8)}{(x-8) x}=10\)

⇒ \(\frac{480 x-480 x+3840}{x^2-8 x}=10\)

⇒ \(10 x^2-80 x=3840\)

⇒ \(x^2-8 x-384=0\)

⇒ \(x^2-24 x+16 x-384=0\)

⇒ \(x(x-24)+16(x-24)=0\)

⇒ (x- 24) (x + 16) =0 ⇒  x = 24    or  x = -16

Since no. of students cannot be negative

Hence, x = 24

No. of students who went for picnic = x- 8 = 24-8=16 students.

Quadratic Equation Exercise 4.1

Question 1. Check whether the following are quadratic equations:

  1. \((x+1)^2=2(x-3)\)
  2. \(x^2-2 x=(-2)(3-x)\)
  3. (x-2)(x+1) =(x-1)(x+3)
  4. \((x-3)(2 x+1)=x(x+5)\)
  5. \((2 x-1)(x-3)=(x+5)(x-1)\)
  6. \(x^2+3 x+1=(x-2)^2\)
  7. \((x+2)^3=2 x\left(x^2-1\right)\)
  8. \(x^3-4 x^2-x+1=(x-2)^3\)

Solution:

1. \((x+1)^2=2(x-3)\)

⇒ x2 + 2x + 1 = 2x- 6

⇒ x2 + 7 =0

The highest power of the variable x in it is 2.

∴ The given equation is a quadratic equation.

2. \(x^2-2 x=(-2)(3-x)\)

x2 -2x =- 6 + 2x

x2-2x-2x+6 =0

x2 -4x + 6 =0

The highest power of the variable x in it is 2.

∴ Given equation is a quadratic equation.

3. (x-2)(x+1) =(x-1)(x+3)

⇒ x(x + 1 ) -2 (x + 1 ) = x(x + 3) – 1 (x + 3)

⇒ x2 +x-2x- 2 =x2 + 3x-x-3

⇒ x2 – x- 2 =x2 + 2x- 3

⇒ x2 +x-2- x2 +3 = 0

⇒ -3x+1 = 0

The highest power of the variable x is not 2, in it.

∴ Given equation is not a quadratic equation.

4. (x-3) (2x+ 1) = x(x + 5)

x(2x+1)-3 (2x + 1) =x2 + 5X

2x2 + x- 6x- 3 – x2 – 5x =0

x2– 10x – 3 =0

The highest power of the variable x in it is 2.

∴ Given equation is a quadratic equation.

5. (2x-1)(x-3)=(x +5)(x-1)

2x(x-3)-1(x-3) = x(x-1) + 5(x- 1)

2x2 – 6x-x + 3 = x2 -x+ 5x- 5

2x2 – 7x + 3 = x2 + 4x- 5

2x2– 7x + 3 -x2 – 4x + 5 = 0

x2– 11x+ 8 = 0

The highest power of variable x is 2 in it.

∴ Given equation is a quadratic equation.

6. x2 + 3x + 1 = (x- 2)2

x2 + 3x + 1 =x2 – 4x + 4

x2 + 3x+ 1 -x2 + 4x- 4 = 0

7x – 3 = 0

The highest power of the variable x is not 2 in it.

∴ Given equation is not a quadratic equation.

7. (x + 2)3 = 2x(x2 – 1 )

⇒ x3 + 3.x.2 .(x + 2) + 23 = 2x3 – 2x

⇒ x3+ 6x2 + 12x + 8 – 2x3 + 2x =0

⇒ -x3+6x2+14x+8 = 0

The highest power of the variable x is not 2 in it.

∴ Given equation is a quadratic equation.

8. \(x^3-4 x^2-x+1=(x-2)^3\)

⇒ \(x^3-4 x^2-x+1=x^3-3 x \cdot 2(x-2)-2^3\)

⇒ \(x^3-4 x^2-x+1=x^3-6 x^2+12 x-8\)

⇒ \(x^3-4 x^2-x+1-x^3+6 x^2-12 x+8=0\)

⇒ \(2 x^2-13 x+9=0\)

Question 2. Represent the following situations in the form of quadratic equations :

  1. The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of tire plot.
  2. The product of two consecutive positive integers is 306. We need to find the integers.
  3. Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
  4. A train travels a distance of 480 1cm at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Solution:

1. Let the breadth of the plot =x meter

∴ Length of plot = (2x + 1) metre

Now, from length x breadth = area

(2x + 1) × x =528

⇒ 2x2 + x =528

⇒ 2x2 + x- 528 =0

which is the required quadratic equation.

2. Let two consecutive positive integers are x and x+ 1

∵ The product of two consecutive positive integers = 306

∴ x(x+1) = 360

⇒ x2+x = 306

⇒ x2+x-306 = 0

which is the required quadratic equation.

3. Let the present age of Rohan = x years

∴ Present age of Rohan’s mother

= (x + 26), years

After 3 years,

Rohan’s age = (x + 3) years

The age of Rohan’s mother = (x + 26 + 3) years

= (x + 29) years

According to the problem,

After 3 years, the product of their ages = 360

(x + 3) (x + 29) = 360

x(x + 29) + 3(x + 29) = 360

x2 + 29x + 3x + 87 = 360

x2 + 32x + 87 – 360 = 0

x2 + 32x- 273 = 0

which is the required quadratic equation.

4. Let the speed of train = A km/hr

Distance = 480 1cm

∴ Time taken to cover a distance of 480 km

⇒ \(\frac{480}{x} \mathrm{hrs}\)

If, the speed of the train = (x- 8) Km/hr  then the time is taken to cover 480 Km distance

⇒ \(\frac{480}{x-8} \mathrm{hrs}\)

According to the problem,

⇒ \(\frac{480}{x-8}-\frac{480}{x}=3\)

⇒ \(\frac{480 x-480(x-8)}{x(x-8)}=3\)

⇒ 480x- 480x + 3840 = 3x(x- 8)

⇒ 3840 = 3(x2 – 8x)

⇒ 1280 =x2– 8x

⇒ 0= x2 -8x -1280

⇒ x2 -8x- 1280 = 0

which is the required quadratic equation.

Quadratic Equation Exercise 4.2

Question 1. Find the roots of the following quadratic equations by factorisation:

  1. x2-3x-10 = 0
  2. 2x2+x-6 = 0
  3. \(\sqrt{2} x^2+7 x+5 \sqrt{2}=0\)
  4. \(2 x^2-x+\frac{1}{8}=0\)
  5. 100x2– 20x + 1=0

Solution:

1. x2-3x-10 = 0

⇒ x2-5 x+2 x-10=0

⇒ x(x-5)+2(x-5)=0

⇒ (x-5)(x+2)=0

⇒ 5=0 or x+2=0

⇒ x=5 or x=-2

∴ Roots of given quadratic equation = 5, -2

2. 2 x^2+x-6=0

⇒ 2 x^2+4 x-3 x-6=0

⇒ 2 x(x+2)-3(x+2)=0

⇒ (x+2)(2 x-3)=0

⇒ x+2 =0 or 2x-3 = 0

⇒ x = -2 or 2x = 3

⇒ x =-2 or x = \(\frac{3}{2}\)

∴ Roots of given quadratic equation = -2, \(\frac{3}{2}\)

3. \(\sqrt{2} x^2+7 x+5 \sqrt{2}=0 \)

⇒ \(\sqrt{2} x^2+5 x+2 x+5 \sqrt{2}=0\)

⇒ \((\sqrt{2} x+5)+\sqrt{2}(\sqrt{2} x+5)=0\)

⇒ \((\sqrt{2} x+5)(x+\sqrt{2})=0\)

⇒ \(\sqrt{2} x+5=0 \quad \text { or } x+\sqrt{2}=0\)

⇒ \(\sqrt{2} x=-5 \text { or }x=-\sqrt{2}\)

⇒ \(x=\frac{-5}{\sqrt{2}} \text { or }x=-\sqrt{2}\)

∴ Roots of given quadratic equation

⇒ \(\frac{-5}{\sqrt{2}},-\sqrt{2} \text {. }\)

4. \(2 x^2-x+\frac{1}{8}=0\)

⇒ \(16 x^2-8 x+1=0\)

⇒ \(16 x^2-4 x-4 x+1=0\)

⇒ \(4 x(4 x-1)-1(4 x-1)=0\)

⇒ (4x-1)(4 x-1)=0

⇒ 4x-1 = 0  4x- 1 = 0

⇒ 4x= 1 or 4x= 1

⇒ \(x=\frac{1}{4} \quad \text { or } \quad x=\frac{1}{4}\)

∴ Roots of given quadratic equation = \(\frac{1}{4}, \frac{1}{4} \text {.}\)

5. \(100 x^2-20 x+1=0\)

⇒ \(100 x^2-10 x-10 x+1=0\)

⇒ 10x(10 x-1)-1(10 x-1)=0

⇒ (10x-1)(10 x-1)=0

⇒ 10x-1=0 or 10 x-1=0

⇒ 10x=1 or 10 x=1

⇒ \(x=\frac{1}{10} \text { or } \quad x=\frac{1}{10}\)

∴ Roots of given quadratic equation \(\frac{1}{10}, \frac{1}{10}\)

Question 2. Represent the following situations mathematically:

  1. John and Jiwanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
  2. A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was 750. We would like to find out the number of toys produced on that day.

Solution : 

1. Let the number of marbles initially with John =x

∴ Initially, the number of marbles with Jiwanti = 45-x

After losing 5 marbles each,

Remaining marbles with John =x- 5

Remaining marbles with Jiwanti = 45 -x- 5

= 40-x

According to the problem,

Product of remaining marbles with them =124

(x- 5) (40-x) = 124

⇒ x(40-x) -5(40-x) = 124

⇒ 40x-x2– 200 + 5x= 124

⇒ -x2 + 45x- 200 = 124

⇒ 0= 124 +x2– 45x + 200

⇒ x2 – 45x + 324 = 0

⇒ x2– 36x -9x + 324 = 0

⇒ x(x- 36) -9(x- 36) = 0

⇒ (x-36) (x-9) = 0

⇒ x-36 = 0 or x-9 = 0

⇒ x = 36 or x = 9

If x = 36 then 45-x = 45-36 = 9

If x = 9 then 45-x = 45-9 = 36

∴ Marbles with John = 36

and marbles with Jiwanti = 9

or

Marbles with John = 9

and marbles with Jiwanti = 36

2. Let the number of toys = x

Cost of each toy =  ₹(55 -x)

Cost of x toys = ₹(55 – x) x-

₹(55x-x2)

According to the problem,

⇒ 55x -x2 = 750

⇒ 0 =x2-55X+750

⇒ x2 – 30x- 25x + 750 =0

⇒ x(x- 30) -25 (x-30) =0

⇒ (x-30) (x-25) =0

⇒ x-30 = 0 or x- 25 = 0

⇒ x = 30 or x = 25

∴ Number of toys produced = 25 or 30.

Question 3. Find two numbers whose sum is 27 and whose product is 182.
Solution:

Let one number = x

∴ Second number =27 -x

According to the problem,

Product of two numbers = 182

⇒ x (27-x) = 182

⇒ 27x -x2 = 182

⇒ 0 =x2 – 27x+ 182

⇒ x2– 13x- 14x+ 182 =0

⇒ x(x- 13) -14 (x- 13) =0

⇒ (x- 13) (x- 14) =0

⇒ x-13 = 0 or x-14 =0

⇒ x= 13 or x = 14

If x = 13, then 27-x = 27-13 = 14

If x = 14, then 27 -14 = 13

Therefore,numbers=(13 and 14)or(14and 13).

Question 4. Find two consecutive positive integers, a sum of whose squares is 365.
Solution:

Given

The sum of whose squares is 365

Let two consecutive positive integers be x and x + 1.

According to the problem,

x2 + (x+1)2 =365

x2 + x2 + 2x + 1 =365

2x2 + 2x + 1 – 365 = 0

2x2 + 2x- 364 = 0

x2+x- 182 =0

x2+14x-13x-182 = 0

=x (x + 14) (x- 13) =0

x + 14=0 or x-13=0

x = -14 or x = 13

x is a positive integer,

∴ neglecting x = -14,

x = 13

x+1 = 13+1 = 14

Therefore, required positive integers = 13 and 14.

Question 5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution:

Given

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm

Let the base of the right triangle = x cm

∴ Its height = (x- 7) cm

Given, the hypotenuse of the right triangle = 13cm

From Pythagoras theorem, in a right triangle (base)2 + (height)2 = (hypotenuse)2

⇒ x2 + (x- 7)2 = 132

⇒ x2+x2– 14x + 49 = 169

⇒ 2x2 – 14x + 49 — 169 =0

⇒ 2x2 -14x- 120 =0

⇒ x2 – 7x- 60 =0

⇒ x2– 12x + 5x -60 =0

⇒ x(x- 12) + 5(x- 12) =0

⇒ (x- 12) (x+ 5) =0

⇒ x-12 =0 or x + 5= 0

⇒ x = 12 or x = -5

but the value of x cannot be negative.

∴ Neglecting

x = -5

x = 12

⇒ x-7 =12-7 = 5

Therefore, the other two sides of the triangle =12 cm and 5 cm.

Question 6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹90, find the number of articles produced and the cost of each article.
Solution:

Given

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹90,

Let the number of pottery articles produced in a day = x

∴ Cost of each article = ₹(2x+ 3)

⇒ Cost of x articles = ₹(2x + 3)x

According to the problem,

(2x + 3) x = 90

⇒ 2x2 + 3x = 90

⇒ 2x2 + 3x-90 =0

⇒ 2x2 + 15x- 12x- 90 = 0

⇒ x(2x+ 15) -6(2x+ 15) =0

⇒ (2x+ 1 5) (x -6) =0

⇒ 2x + 15 = 0 or x -6 = 0

⇒ \(x=-\frac{15}{2} \text { or } \quad x=6\)

but x = \(-\frac{15}{2}\) is not possible

∴ x = 6

⇒ 2x + 3 = 2×6 + 3=15

Therefore, the number of pottery articles in a day = 6, and the cost of each article =₹15.

Quadratic Equation Exercise 4.3

Question 1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square :

  1. 2x2 – 7x + 3 = 0
  2. 2x2 + x- 4 = 0
  3. \(4 x^2+4 \sqrt{3} x+3=0\)
  4. 2x2 +x + 4 = 0

Solution:

1. 2x2 – 7x + 3 = 0

⇒ \(x^2-\frac{7}{2} x+\frac{3}{2}=0\)

⇒ \( {\left[x^2-2 x \cdot \frac{7}{4}+\left(\frac{7}{4}\right)^2\right]+\frac{3}{2}-\left(\frac{7}{4}\right)^2 }=0\)

⇒ \(\left(x-\frac{7}{4}\right)^2+\frac{24-49}{16}=0\)

⇒ \(\left(x-\frac{7}{4}\right)^2-\frac{25}{16}=0\)

⇒ \(\left(x-\frac{7}{4}\right)^2-\left(\frac{5}{4}\right)^2=0\)

⇒ \(\left(x-\frac{7}{4}-\frac{5}{4}\right)\left(x-\frac{7}{4}+\frac{5}{4}\right)=0\)

⇒ \((x-3)\left(x-\frac{1}{2}\right)=0\)

⇒ \(x-3=0\text { or }x-\frac{1}{2}=0\)

⇒ \(x=3\text { or }x=\frac{1}{2}\)

∴ Roots of given equation = \(3, \frac{1}{2}\)

2. \(2 x^2+x-4=0\)

⇒ \(x^2+\frac{1}{2} x-2=0\)

⇒ \(x^2+2 x \cdot \frac{1}{4}+\left(\frac{1}{4}\right)^2-2-\left(\frac{1}{4}\right)^2=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2-\left(\frac{32+1}{16}\right)=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2-\frac{33}{16}=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2-\left(\frac{\sqrt{33}}{4}\right)^2=0\)

⇒ \(\left(x+\frac{1}{4}-\frac{\sqrt{33}}{4}\right)\left(x+\frac{1}{4}+\frac{\sqrt{33}}{4}\right)=0\)

⇒ \(x+\frac{1}{4}-\frac{\sqrt{33}}{4}=0 \text { or } x+\frac{1}{4}+\frac{\sqrt{33}}{4}=0\)

⇒ \(x=\frac{\sqrt{33}}{4}-\frac{1}{4} \quad \text { or } \quad x=-\frac{\sqrt{33}}{4}-\frac{1}{4}\)

⇒ \(x=\frac{\sqrt{33}-1}{4} \quad \text { or } \quad x=-\left(\frac{\sqrt{33}+1}{4}\right)\)

∴ Roots of a given equation

⇒ \(\frac{\sqrt{33}-1}{4},-\left(\frac{\sqrt{33}+1}{4}\right)\)

3. \(4 x^2+4 \sqrt{3} x+3=0\)

⇒ \(x^2+\sqrt{3} x+\frac{3}{4}=0\)

⇒ \(x^2+2 x \cdot \frac{\sqrt{3}}{2}+\left(\frac{\sqrt{3}}{2}\right)^2=0\)

⇒ \(\left(x+\frac{\sqrt{3}}{2}\right)^2=0\)

⇒ \(x+\frac{\sqrt{3}}{2}=0 \text { or }x+\frac{\sqrt{3}}{2}=0\)

⇒ \(x=-\frac{\sqrt{3}}{2}\text { or }x=-\frac{\sqrt{3}}{2}\)

∴ Roots of given equation = \(-\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}\)

4. \(2 x^2+x+4=0\)

⇒ \(x^2+\frac{1}{2} x+2=0\)

⇒ \(x^2+2 x \cdot \frac{1}{4}+\left(\frac{1}{4}\right)^2+2-\left(\frac{1}{4}\right)^2=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2+\frac{32-1}{16}=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2=\frac{-31}{16}\)

⇒ \(x+\frac{1}{4}=\sqrt{-\frac{31}{16}}\) which is an imaginary number.

∴ Roots of given equation does not exist.

Question 2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
Solution:

1. 2x2 – 7x + 3 = 0

On comparing with ax2 +bx + c,

a = 2,b=-7,c = 3

Now, \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-(-7) \pm \sqrt{(-7)^2-4(2)(3)}}{2(2)}\)

⇒ \(\frac{7 \pm \sqrt{49-24}}{4}=\frac{7 \pm \sqrt{25}}{4}=\frac{7 \pm 5}{4}\)

⇒ \(x=\frac{7+5}{4}\text { or }x=\frac{7-5}{4}\)

⇒ \(x=3\text { or }x=\frac{1}{2}\)

∴ Roots of equation = \(3, \frac{1}{2}\)

2. 2x2 +x- 4 = 0

On comparing with ax2 + bx + c = 0

a = 2,b = 1,c = -4

Now, \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-1 \pm \sqrt{(1)^2-4(2)(-4)}}{2 \times 2}=\frac{-1 \pm \sqrt{1+32}}{4}\)

⇒ \(\frac{-1 \pm \sqrt{33}}{4}\)

⇒ \(x=\frac{-1+\sqrt{33}}{4} \text { or } \frac{-1-\sqrt{33}}{4}\)

∴ Roots of equation = \(\frac{-1+\sqrt{33}}{4}, \frac{-1-\sqrt{33}}{4}\)

3. \(4 x^2+4 \sqrt{3} x+3=0\)

On comparing with ax2 +bx + c = 0

a = 4, b = \(4 \sqrt{3}\), c = 3

Now, \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-4 \sqrt{3} \pm \sqrt{(4 \sqrt{3})^2-4 \times 4 \times 3}}{2 \times 4}\)

⇒ \(\frac{-4 \sqrt{3} \pm \sqrt{48-48}}{8}=\frac{-4 \sqrt{3} \pm \sqrt{0}}{8}\)

⇒ \(-\frac{4 \sqrt{3}}{8}=-\frac{\sqrt{3}}{2}\)

Number of roots of a quadratic equation = 2

∴ Roots of given equation = \(-\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}\)

4. 2x2 + x + 4 = 0

On comparing with ax2, + bx + c = 0

a = 2,b=1,c = 4

Now, \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-1 \pm \sqrt{(1)^2-4(2)(4)}}{2 \times 2}\)

⇒ \(\frac{-1 \pm \sqrt{1-32}}{4}=\frac{-1 \pm \sqrt{-31}}{4}\)

⇒ \(\sqrt{-31}\)  is an imaginary number,

∴ the values for are imaginary.

So, the real roots of the given equation do not exist.

Question 3. Find the roots of the following equations:

  1. \(x-\frac{1}{x}=3, x \neq 0\)
  2. \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}, x \neq-4,7\)

Solution:

1. \(x-\frac{1}{x}=3, x \neq 0\)

⇒ \(\frac{x^2-1}{x}=3\)

⇒ \(x^2-1=3 x \Rightarrow x^2-3 x-1=0\)

On comparing with ax2 + bx + c = 0

a= 1, b = -3, c =-1

x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-1)}}{2(1)}=\frac{3 \pm \sqrt{9+4}}{2}\)

⇒ \(\frac{3 \pm \sqrt{13}}{2}\)

⇒ \(x=\frac{3+\sqrt{13}}{2} \text { or } \quad x=\frac{3-\sqrt{13}}{2}\)

Therefore, the roots of given equations

⇒ \(\frac{3+\sqrt{13}}{2}, \frac{3-\sqrt{13}}{2} \text {. }\)

2. \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}, x \neq-4,7\)

⇒ \(\frac{(x-7)-(x+4)}{(x+4)(x-7)}=\frac{11}{30}\)

⇒ \(\frac{x-7-x-4}{x(x-7)+4(x-7)}=\frac{11}{30}\)

⇒ \(\frac{-11}{x^2-7 x+4 x-28}=\frac{11}{30}\)

⇒ \(\frac{-1}{x^2-3 x-28}=\frac{1}{30}\)

⇒ \(x^2-3 x-28=-30\)

⇒ \(x^2-3 x-28+30=0\)

⇒ \(x^2-3 x+2=0\)

⇒ \(x^2-2 x-x+2=0\)

⇒ x(x-2)-1(x-2)=0

⇒ (x-2)(x-1)=0

⇒ x-2 = 0 or x-1 = 0

⇒ x = 2 or x = 1

∴ Roots of given equation = 2,1

Question 4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\). Find his present age.
Solution:

Given

The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\).

Let the present age of Rehman = x years

3 years before, Rehman’s age = (x- 3) years

After 5 years, Rehman’s age = (x + 5) years

According to the problem, \(\frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}\)

⇒ \(\frac{(x+5)+(x-3)}{(x-3)(x+5)}=\frac{1}{3}\)

⇒ \(\frac{2 x+2}{x(x+5)-3(x+5)}=\frac{1}{3}\)

⇒ \(\frac{2 x+2}{x^2+5 x-3 x-15}=\frac{1}{3}\)

⇒ \(\frac{2 x+2}{x^2+2 x-15}=\frac{1}{3}\)

⇒ \(x^2+2 x-15=6 x+6\)

⇒ \(x^2+2 x-15-6 x-6=0\)

⇒ \(x^2-4 x-21=0\)

⇒ \(x^2-7 x+3 x-21=0\)

⇒ x(x-7)+3(x-7)=0

⇒ (x-7)(x+3)=0

⇒ x-7=0 or  x+3=0

⇒ x=7 or  x=-3

but the age cannot be negative.

∴ x = 7

⇒ Rehman’s age = 7 years.

Question 5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Solution:

Given

In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210.

Let marks obtained by Shefali in Mathematics =x

∴ Her marks in English = (30- x)

If, marks in Mathematics = x + 2

marks in English =30-x-3 = 27-x

then (x + 2) (27 -x) =210

⇒ x(27 -x) + 2(27 -x) =210

⇒ 27x -x2 + 54 – 2x = 210

⇒ 25x-x2 + 54 =210

⇒ 0 = x2– 25x- 54 + 210

⇒ x2 – 25x + 156 =0

⇒ x2– 13x- 12x+ 156 =0

⇒ x(x -13) – 12(x – 13) =0

⇒ (x- 13) (x- 12) =0

⇒ x -13 = 0 or x-12=0

⇒ x -13 or x-12

If x = 13 then 30-x = 30-13 = 17

If x = 12 then 30-x = 30-12 = 18

So, for Shefali

Maries in Mathematics = 13

and marks in English = 17

or

Maries in Mathematics = 12

and marks in English = 18

Question 6. The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.
Solution:

Given

The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side,

Let the smaller side of the rectangular field =xmetre = breadth

Larger side = (x + 30) metre = length

diagonal = (x + 60) metre

Now, (length) + (breadth) = (diagonal)

⇒ x2 + (x+ 30)2 = (x+60)2

⇒ x2 +x2 + 60x + 900 =x2 + 120x + 3600

⇒ x2 +x2 + 60x + 900 -x2 – 120x- 3600 = 0

⇒ x2– 60x -2700 = 0

⇒ x2– 90x + 3x -2700 =0

⇒ x(x- 90) + 30(x- 90) = 0

⇒ (x-90) (x + 30) =0

⇒ x-90 =0 or x+30 = 0

⇒ x = 90 or x = -30

but the side cannot be negative.

⇒ x = 90

⇒ x+30 = 90+30 = 120

Therefore, sides of the rectangular field = 120 m and 90 m.

Question 7. The difference of the squares of two numbers is 1 80. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution:

Given

The difference of the squares of two numbers is 1 80. The square of the smaller number is 8 times the larger number.

Let smaller number =x

∴ Larger number × 8 = x2

⇒ Larger number = \(\frac{x^2}{8}\)

According to the problem, (larger number)2 – (smaller number)2 =180

⇒ \(\left(\frac{x^2}{8}\right)^2-x^2=180\)

⇒ \(\left(\frac{y}{8}\right)^2-y=180\)

⇒ \(\text { where } x^2=y \text { (say) }\)

⇒ \(\frac{y^2}{64}-y=180\)

⇒ \(y^2-64 y=11\)

⇒ \(y^2-64 y-11520=0\)

⇒ \(y^2-144 y+80 y-11520=0\)

⇒ \(y(y-144)+80(y-144)=0\)

⇒ \((y-144)(y+80)=0\)

⇒ \(y-144=0 \text { or } y+80=0\)

⇒ \(y=144 \text { or } y=-80\)

⇒ \(x^2=144 \text { or } x^2=-80\)

x2 =- 80 is not possible.

∴ \(x^2=144\)

⇒ \(x= \pm 12\)

⇒ \(x=12 \quad \text { or } \quad x=-12\)

⇒ \(\frac{x^2}{8}=\frac{144}{8}=18\)

Therefore, a number are 12, 18 or -12, 18.

Question 8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:

Given

A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey.

Let the speed of the train =x km/hr

∴ Time taken to cover 360 km distance = \(\frac{360}{x} \mathrm{~km}\)

If the speed of train = (x+ 5) km/hr

then time taken to cover 360 km distance = \(\frac{360}{x+5} \mathrm{hr}\)

According to the problem, \(\frac{360}{x}-\frac{360}{x+5}=1\)

⇒ \(\frac{360(x+5)-360 x}{x(x+5)}=1\)

⇒ \(\frac{360 x+1800-360 x}{x^2+5 x}=1\)

⇒ \(\frac{1800}{x^2+5 x}=1\)

⇒ \(x^2+5 x=1800\)

⇒ \(x^2+5 x-1800=0\)

⇒ \(x^2+45 x-40 x-1800=0\)

⇒ x(x+45)-40(x+45)=0

⇒ (x+45)(x-40)=0

⇒ x+45=0  or x-40=0

⇒ x=-45  or  x=40

but speed cannot be negative.

∴ Speed of train = 40 km/hr.

Question 9. Two water taps together can fill a tank in \(9 \frac{3}{8}\)hours. The tap of a larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:

Given

Two water taps together can fill a tank in \(9 \frac{3}{8}\)hours. The tap of a larger diameter takes 10 hours less than the smaller one to fill the tank separately.

Let the time taken to fill the tank by the tap of smaller diameter =x hours

Time taken to fill the tank by the tap of larger diameter = (x- 10) hours

Now, work done by tap of smaller diameter in 1 hour = \(\frac{1}{x}\)

and work done by tap of larger diameter in 1hour = \(\frac{1}{x}+\frac{1}{x-10}\)

Work done by two taps in 1 hour = \(\frac{1}{x}+\frac{1}{x-10}\)

According to the problem, both taps fill the Time taken by the passenger train to cover 132 1cm tank in \(9 \frac{3}{8}=\frac{75}{8} \)

∴ \(\left(\frac{1}{x}+\frac{1}{x-10}\right) \times \frac{75}{8}=1\)

⇒ \(\frac{1}{x}+\frac{1}{x-10}=\frac{8}{75}\)

⇒ \(\frac{x-10+x}{x(x-10)}=\frac{8}{75}\)

⇒ \(\frac{2 x-10}{x^2-10 x}=\frac{8}{75}\)

⇒ \(8 x^2-80 x=150 x-750\)

⇒ \(8 x^2-80 x-150 x+750=0\)

⇒ \(8 x^2-230 x+750=0\)

⇒ \(4 x^2-115 x+375=0\)

⇒ \(4 x^2-100 x-15 x+375=0\)

⇒ 4 x(x-25)-15(x-25)=0

⇒ (x-25)(4 x-15)=0

⇒ x-25=0  or  4 x-15=0

⇒ x-25 or \(x=\frac{15}{4}\)

but \(x=\frac{15}{4}\) is not possible because both taps fill the tank in \(9 \frac{3}{4}\) hours.

⇒ x = 25 and x- 10 = 25 – 10= 15

Therefore, the tap with a smaller diameter fills the tank in 25 hours and the tap with a larger diameter fills the tank in 15 hours.

Question 10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Solution:

Given

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train

Let the average speed of passenger train =x km/hr.

∴ Average speed of express train = (x+11) km/hr

Time taken by passenger train to cover 1321cm distance = \(\frac{132}{x} \mathrm{hr}\)

and time taken by express train to cover 1321cm distance = \(\frac{132}{x+11} \mathrm{hr}\)

According to the problem,

⇒ \(\frac{132}{x}-\frac{132}{x+11}=1\)

⇒ \(\frac{132(x+11)-132 x}{x(x+11)}=1\)

⇒ \(\frac{132 x+1452-132 x}{x^2+11 x}=1\)

⇒ \(x^2+11 x=1452\)

⇒ \(x^2+11 x-1452=0\)

⇒ \(x^2+44 x-33 x-1452=0\)

⇒ x(x+44)-33(x+44)=0

⇒ (x+44)(x-33)=0

⇒ x+44=0  or  x-33=0

⇒ x=-44  or  x=33

but the speed cannot be negative.

∴ x = 33

⇒ x + 11 =33+ 11 =44

Average speed of passenger train = 33 km/hr

an average speed of express train = 44 km/hr

Question 11. The sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:

Given

The sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m

Let side of a square = x metre

Perimeter of this square = 4x metre

∴ Perimeter of second square = (4x + 24) metre

⇒ Side of second square = \(\frac{4 x+24}{4}\) = (x + 6) metre

Now, area of first square =x2 m2 and area of second square = (x+ 6)2 m2

According to the problem,

⇒ \((x+6)^2+x^2=468\)

⇒ \(x^2+12 x+36+x^2-468=0\)

⇒ \(2 x^2+12 x-432=0\)

⇒ \(x^2+6 x-216=0\)

⇒ \(x^2+18 x-12 x-216=0\)

⇒ x(x+18)-12(x+18)=0

⇒ (x+18)(x-12)=0

⇒ x+18=0   or   x-12=0

⇒ x=-18   or x=12

but the side of a square cannot be negative.

∴ x= 12

⇒ x+ 6 = 12 + 6= 18

Therefore, sides of squares = 12 m and 18 m

Quadratic Equation Exercise 4.4

Question 1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them :

  1. \(2 x^2-3 x+5=0\)
  2. \(3 x^2-4 \sqrt{3} x+4=0\)
  3. \(2 x^2-6 x+3=0\)

Solution:

1. \(2 x^2-3 x+5=0\)

On comparing with ax2 + bx + c = 0

a = 2, b=-3,c = 5

Discriminant D = b2 – 4ac = (-3)2 – 4(2) (5)

= 9-40 =-31

∵ D is negative

∴ Roots of the equation are imaginary.

Here, the real roots of the equation do not exist.

2. \(3 x^2-4 \sqrt{3} x+4=0\)

On comparing with ax2 + bx + c = 0

a = 3,b =\(-4 \sqrt{3}\), c= 4

Discriminant D = b2– 4ac

⇒ \((-4 \sqrt{3})^2-4(3)(4)\)

= 48-48 = 0

The roots of given equation are real and equal

Now, \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{4 \sqrt{3} \pm 0}{2 \times 3}=\frac{2}{\sqrt{3}}\)

Roots of given equation = \(\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)

3. 2x2 – 6x + 3=0

On comparing with ax2 + bx + c = 0

a = 2, b = – 6, c = 3

∴ Discriminant D = b2– 4ac = (-6)2– 4 (2) (3)

= 36-24 = 12 > 0

∵ D is positive

Roots of the given equation are real and unequal

Now, \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{6 \pm \sqrt{12}}{2 \times 2}=\frac{6 \pm 2 \sqrt{3}}{4}\)

⇒ \(x=\frac{3 \pm \sqrt{3}}{2}\)

⇒ \(\frac{3+\sqrt{3}}{2} \text { or } \frac{3-\sqrt{3}}{2} \text {. }\)

Question 2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.

  1. 2x2+kx+ 3 = 0
  2. k x(x- 2) + 6 = 0

Solution:

2x2+kx+ 3 = 0

On comparing with ax2 +bx + c = 0

a = 2, b=k, c = 3

DiscriminantD = b2– 4ac = k2 – 4 x 2 x 3

= k2 -24

Given that, the roots are equal.

∴ D = 0  ⇒ k2 – 24 = 0

⇒ k2 = 24

⇒ K = \(\pm \sqrt{24}= \pm 2 \sqrt{6}\)

2.  kx(x -2) + 6 = 0

kx2 – 2kx+6 = 0

On comparing with a2 + bx + c = 0

a = k, b = -2k, c = 6

∴ Discriminant D = b2– 4ac = (-2k)2 – 4k(6)

= 4k2– 24k = 4k(k- 6)

Given that the roots are equal.

∴ D = 0

⇒ 4k(k – 6) = 0

⇒ k – 6 = 0

⇒ k = 6

Question 3. Is It possible to design a rectangular mango grove whose length Is twice its breadth, and the area is 800 m2? If so, find its length and breadth.
Solution:

Let the breadth of mango grove

= x metre

∴ Length = 2x metre

Area of grove = length x breadth

= (2x)(x) = 2x2

According to the problem,

⇒ 2x2 = 800

⇒ x2 = 400

⇒ x = ± 20

but the breadth cannot be negative.

∴ x = 20

2x = 2 × 20 = 40

Therefore, a grove is possible, and its length = 40m and breadth = 20 m.

Question 4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their age in years was 48.
Solution:

Let the present age of one friend =  x years

∴ Present age of second friend = (20- x) years

4 years ago, the age of first friend = (x- 4) years

The age of second friend = 20 – x- 4

= (16- x) years

According to the problem,

(x-4) (16 -x) =48

⇒ x(1 6 – 4) -4(16 – x) =48

⇒ 16x – x2– 64 + 4x = 48

⇒ 20x-x2 -64 = 48

⇒ 0 = 48 – 20x + x2 + 64

x2-20x+ 112=0 ….(1)

On comparing with ax2 + bx + c = 0

a = 1, b = -20, c = 112

Now, discriminant D=b2 – 4ac

= (- 20)2 – 4 (1) (112)

= 400 -448 =-48 <0

Roots of equation (1) are imaginary.

Therefore, the given condition is not possible.

Question 5. Is it possible to design a rectangular park of perimeter 80 m and an area of 400 m2? If so, find its length and breadth.
Solution:

Let the length of the park = x meter

Now, 2 (length + breadth) = perimeter

2(x + breadth) = 80 metre

x+ breadth = 40 metre

breadth = (40 – x) metre

According to the problem,

Area of park= 400 m2

x(40 – x) = 400

40x-x2 = 400

0 = x2 – 40x + 400

x2 – 40x + 400 = 0

On comparing with ax2 +bx+ c = 0

a = 1, b = -40, c = 400

Discriminant D = b2 – 4ac

= (-40)2 – 4(1)(400)

= 1600- 1600 = 0

⇒ Roots are equal

Now, \(x=\frac{-b \pm \sqrt{D}}{2 a}=\frac{40 \pm 0}{2 \times 1}=20\)

So, such a park is possible.

Its length = breadth = 20m.

Quadratic Equation Multiple Choice Questions

Question 1. Which of die following equations has the root 3?

  1. \(x^2+x+1\)
  2. \(x^2-4 x+3=0\)
  3. \(3 x^2+x-1=0\)
  4. \(x^2+9=0\)

Answer: 2. \(x^2-4 x+3=0\)

Question 2. The sum of roots of the equation 5x2 – 3x + 2 = 0 is

  1. \(\frac{3}{5}\)
  2. \(-\frac{3}{5}\)
  3. \(\frac{2}{5}\)
  4. \(-\frac{2}{5}\)

Answer: 1. \(\frac{3}{5}\)

Question 3. The quadratic equation \(2 x^2-\sqrt{5} x+1=0\) has

  1. Two distinct real roots
  2. Two equal roots
  3. No real root
  4. More than two real roots

Answer: 3. No real root

Question 4. One root of the equation x2 + k x + 4 = 0 is -2. The value of k is :

  1. -2
  2. 2
  3. -4
  4. 4

Answer: 4. 4

Question 5. The value of k for which die roots of equation 2kx2 – 6x + 1 = 0 is equal, is :

  1. \(-\frac{9}{2}\)
  2. \(\frac{9}{2}\)
  3. 9
  4. -9

Answer: 2. \(\frac{9}{2}\)

Question 6. Which of the following is a quadratic equation?

  1. (x + 2)2 =x2– 5x + 3
  2. x3 +x2 = (x- 1)3
  3. 3x2 + 1 = (3x- 2) (x + 5)
  4. 5x- 7 = 1 + x

Answer: 2. x3 +x2 = (x- 1)3

Question 7. If the product of roots of the equation 5×2- 3x + /{ = 0 is 2, then the value of k is:

  1. 1
  2. 2
  3. 5
  4. 10

Answer: 4. 10

Question 8. The discriminant of quadratic equadon 3xx3 +x2 = (x- 1)3– 6x + 4 = 0 is :

  1. 12
  2. 13
  3. -12
  4. \(3 \sqrt{6}\)

Answer: 3. -12

Question 9. The roots of the quadratic equation x2– 4 = 0 are :

  1. ± 0.2
  2. ±1
  3. ± 2
  4. ±4

Answer: 3. ± 2

Question 10. The discriminant of equation \(3 x^2-2 x+\frac{1}{3}=0\) will be:

  1. 3
  2. 2
  3. 1
  4. 0

Answer: 4. 0

Question 11. If the roots of the quadratic equation 3x2 – 12x + m = 0 are equal, then the value of m will be :

  1. 4
  2. 7
  3. 9
  4. 12

Answer: 4. 12

NCERT Exemplar Solutions for Class 10 Maths Chapter 2 Polynomials

Polynomials

An Expression of the form p(x)= a0+a0x+a0x0…..+a0xn

For example.,

5x+1 is a polynomial of degree in x. Here, a0,a1,a2,…..an are real numbers.

2x2-x-1 is apolynomial in x of degree 2.

y3-2y2+y+5 is a polynomial in y of degree 3.

5z4-3z+1 is a polynomial in z of degree 4.

Constant Polynomial

A polynomial of degree zero is called a constant polynomial.

For example., p(x) = -3, q(x) = 2,f(x) = \(\sqrt{2}\) etc., are constant polynomials. (These are independent of variable)

Zero Polynomial

It is also a constant polynomial with a particular constant value of 0.

So,f(x) = 0 is a zero polynomial. Its degree is not defined, as we cannot say definitely about its degree. Forms of zero polynomial may be

f(x) = 0 = 0 . x2 – 0 .x + 0 = 0 . x5 – 0 . x2= 0 . J-5 + 0. x9 + 0. x4 – 0 . x2 + 0., etc.

So, we cannot say anything about the degree of a zero polynomial.

Linear Polynomial

A polynomial of degree 1 is called a linear polynomial.

It is of the form

p(x) = ax + b,

where a ≠ 0

For example., 3x + 5, 5 – 2x, etc.

Quadratic Polynomial

A polynomial of degree 2 is called a quadratic polynomial. It is of the form

p(x) = ax2 +bx + c, where a ≠ 0

x2 + 5x + 1, 3x2 – x + 1, 5 – x2 etc.

NCERT Exemplar Solutions For Class 10 Maths Chapter 2 Polynomials

Cubic Polynomial

A polynomial of degree 3 is called a cubic polynomial.

It is of the form

p(x) = ax3 +bx2 +cx + d, where a ≠ 0

For example., x3 – x + 1, 5x2 – 4x2 – 2x + 1 etc.

Bi-quadratic Polynomial (or Quartic)

A polynomial of degree 4 is called a biquadratic polynomial.

It is of the form

p(x) = ax4 + bx3 + cx2 + dx + e, a ≠ 0

For example., 2x4 -x2 + 1, 2- 3x +x2 + 4x2 -x4 etc.

Note

Some other names for polynomials are:

Polynomials Some Other Names Of Polynomials Are

Value Of A Polynomial At A Given Point

Let p(x) be a polynomial in x and a is any real number. Then the value obtained by putting x = a in the polynomial p(x) is called the value of the polynomial p(x) at x = a.

This value is denoted by p(a).

P(x) = x2 +x- 1,

then p(5) = 52+5-1= 25+5 = 29

Zeroes Of A Polynomial

A really a is called a zero of a polynomial P(x), if P(α) = 0

or, in other words,

zero of a polynomial is the real value of a variable that vanishes the whole polynomial i.e., the value of a variable that makes the whole polynomial zero.

For example., if P(x) = x2-3x+2

p(2) = 22-3×2+2 = 4-6+2 = 0

∴ 2 is a zero of p(x)

To find the zero/es of a polynomial p(x), put the polynomial equation p(x) = 0.

For example,

Find the zeroes of polynomial p(x) = x2 – 5x – 6

We know that zero of a polynomial is the value of a variable by which p(x) = 0.

⇒ x2 – 5x – 6 = 0 ⇒ (x – 6) (x + 1) = 0

∴ either x – 6 = 0 or x+1=0

⇒ x = 6 or x= -1

So, 6 and -1 are two values of x which make the value of polynomial zero. (You can check on putting these values directly They will make p(x) = 0

Hence, 6 and -1 are the zeroes of the given polynomial.

2. Find the zeroes of polynomial p(x) = x2 – 4x + 4

For zeroes, p(x) = 0

⇒ x2-4x +4 = 0

⇒ (x – 2)2 = 0x – 2 = 0

⇒ x = 2

or x2-4x + 4 = 0 ⇒ (x – 2)(x – 2) = 0

⇒ x = 2, 2

It means, in this case, we get the repeated zeroes. But, we shall say the zero of this polynomial is 2 not the 2 and 2. To say 2 and 2 has no sense. So, if a polynomial has 2 or 3 or more repeated zeroes, we will say only one zero it has.

3. Find the zeros of polynomial p(x)_ = ax2+bx+c, a≠0

To find the zeroes, we put p(x) = 0

∴ \(a x^2+b x+c=0\)

⇒ \(x^2+\frac{b x}{a}+\frac{c}{a}=0\)  (dividing both sides by a)

⇒ \(x^2+\frac{b}{a} x+\underbrace{\frac{b^2}{4 a^2}-\frac{b^2}{4 a^2}}_{\begin{array}{c}
\text { Adding and subtracting } \\
\text { the same quantity }
\end{array}}+\frac{c}{a}=0\)

⇒ \(\left[\text { adding and subtracting }\left(\frac{\text { coff. of } x}{2}\right)^2\right]\)

⇒ \( \left(x+\frac{b}{2 a}\right)^2-\left(\frac{b^2}{4 a^2}-\frac{c}{a}\right)=0 \)

⇒ \(\left(x+\frac{b}{2 a}\right)^2-\left(\frac{b^2-4 a c}{4 a^2}\right)=0\)

⇒ \(\left(x+\frac{b}{2 a}\right)^2-\left(\frac{\sqrt{b^2-4 a c}}{2 a}\right)^2=0\)

(every number is the square of its square root)

⇒ \(\left(x+\frac{b}{2 a}+\frac{\sqrt{b^2-4 a c}}{2 a}\right)\left(x+\frac{b}{2 a}-\frac{\sqrt{b^2-4 a c}}{2 a}\right)=0\left[a^2-b^2=(a+b)(a-b)\right]\)

⇒ \(x=\frac{-b-\sqrt{b^2-4 a c}}{2 a} \text { or } x=\frac{-b+\sqrt{b^2-4 a c}}{2 a}\)

There are two zeroes of a given polynomial.

From the above 3 examples, we observe some very special points :

Zeroes are the values of x when p(x) orj; = 0. It means for example (6, 0) and (-1, 0) must be the points on the polynomial as well as on the x-axis (because the y-coordinate is zero) i.e., the intersection point of the polynomial and x-axis.

Conversely, we can say that if we draw the graph of a polynomial then the x-coordinate or abscissa of the intersection point of the curve and x-axis will give us the zeroes of the polynomial.

From the example, we notice that repeated zeroes will occur only when the polynomial is a perfect square. We still say it as only one zero. We also get only one zero from a linear polynomial, then what is the difference between linear polynomial and quadratic polynomial when we get only one zero in both cases?

So, our conclusion in this case is that a linear polynomial will be a straight line that will cut the x-axis at one point, and the x-coordinate of that point is the zero of that polynomial while a quadratic polynomial in this case of repeated zeroes will not cut the x-axis but it will touch the x-axis and turns thereafter. This is the basic difference between both types of polynomials in the case of one zero.

For example, we see that if x = 6 is a zero of the polynomial then, (x – 6) is one factor of the polynomial. If x = -1 is a zero of the polynomial then (x+ 1) is also one factor of polynomial.

So, if x = α and x = β are the zeroes of a polynomial then, necessarily (x – α) and (x – β) are the factors of that polynomial.

So, if α and β are the zeroes of a polynomial then the polynomial will be In the form (x-α) (x-β). Is it correct 100%? Perhaps, not, Why?

If we find the zeroes of x2-5x-6, we get 6 and -1 as zeros, if we find the zeros of 2x2-10x-12 i.e., 2(x2-5x-6) we also get 6 and -1 as zeros. If we find zeros of 3x2-15x-18 i.e., 3(x2-5x-6). we also get 6 and -1 as the zeroes. If we find the zero of

⇒ \(-\frac{1}{3} x^2+\frac{5}{3} x+2 \text { i.e… }-\frac{1}{3}\left(x^2-5 x-6\right)\)

also get the zeroes as 6 and -1.

Then, if it is given that a and |3 arc the zeroes of a quadratic polynomial then the quadratic polynomial will be k (x – α) (x – β), where k may be any non-zero real number.

Now, let us study the curves (polynomials) graphically.

Geometrical Meaning Of Zeroes Of A Polynomial

Case 1: Consider the first-degree polynomial p(x) = ax + b, a≠0. We know that the graph of y = ax + b is a straight line for example., consider the equation y = 2x – 3. This line passes through the point (1, -1) and (2, 1). Draw the graph of this line as shown. It crosses the x-axis at a point this linear equation has exactly one zero namely \(\frac{3}{2}\).

Thus, for the polynomial p(x) = ax + b, (a≠0) zero of p(x) is \(x=-\frac{b}{a}\)

Case 2: Consider the second-degree polynomial p(x) = ax + bx + c, a ≠ 0.

Consider an example say p(x) = x2 – 5x + 6. Let us see the graph of y = x2 -5x + 6.

Polynomials Consider The Second Degree Polynomial

Draw the graph of this line as shown.

Polynomials The First Degree Polynomial

This graph intersects the j-axis at two points namely (2, 0) and (3, 0). In fact the graph of y = ax2 + bx + c, a ≠ 0 has one of the two shapes either open upwards like when a > 0 or open downward like A when a < 0. These curves are called parabolas.

Polynomials The Second Degree Polynomial

1. When the graph cuts the x-axis at two points A and A’.

Polynomials X Axis At Two Points

Here, the x coordinates of A and A’ are two zeroes of the quadratic polynomial.

2. When the graph cuts the x-axis at one point i.e., at two coincident points.

Polynomials X Axis At One Point

The x coordinate of A is the only zero of the quadratic polynomial.

3. When the graph is either completely above the x-axis or completely below the x-axis i.e., it does not cut the x-axis at any point.

Polynomials X Axis At Any Point

The quadratic polynomial has no zero in this case.

Case 3: Consider the third-degree polynomial p(x) = ax3 + bx2 + cx + d, a ≠ 0.

It can have at most 3

zeroes, depending upon the point of intersection of y = ax3 +bx2+cx+d and x-axis.

Relation Between Zeroes And Coefficients Of A Quadratic Polynomial

Consider the quadratic polynomial

p(x) – ax2 +bx + c, a ≠ 0

Let α, and β be the zeroes of p(x), then (x – α) and (x – β) will be the factors of p(x).

ax2 + bx + c = k(x – α) (x – β), k ≠ 0

= k [x2 – (α + β)x + αβ]

= kx2 – k(α + β)x + Kαβ

On comparing the coefficients of like powers on both sides, we get

k = a

⇒ \(-k(\alpha+\beta)=b \Rightarrow-a(\alpha+\beta)=b \Rightarrow \alpha+\beta=\frac{-b}{a}\)

⇒ \(k \alpha \beta=c \Rightarrow \quad a \alpha \beta=c \Rightarrow \alpha \beta=\frac{c}{a}\)

∴ \(\text { sum of zeroes }=-\frac{\text { coefficient of } x}{\text { coefficient of } x^2},\)

⇒ \(\text { product of zeroes }=\frac{\text { constant term }}{\text { coefficient of } x^2}\)

If a and p are the zeroes of a quadratic polynomial p(x), then

p(x) = k[x2 – (α + β)x + αβ], k ≠ 0

Quadratic Polynomial Solved Examples

Question 1. Find zeroes of the polynomial x – 3x + 2 and verify the relation between its zeroes and coefficients.
Solution:

Given x – 3x + 2

Let p(x) =x2-3x + 2 = x2-2x-x + 2

= x(x-2)-1 (x-2) = (x-2) (x- 1)

∴ P(x) = 0

⇒ (x- 2)(x – 1) = 0

⇒ x – 2 = 0 or x – 1 = 0

⇒ x = 2 or x – 1

∴ Zeroes of p(x) are 2 and 1

Now, Sum of zeroes = 2 + 1 = 3 = \(-\frac{-3}{1}=-\frac{\text { coefficient of } x}{\text { coefficient of } x^2}\)

and product of zeroes = (2) (1) = 2 = \(\frac{2}{1}=\frac{\text { constant term }}{\text { coefficient of } x^2}\)

Question 2. Find zeroes of the quadratic \(\sqrt{3} x^2-8 x+4 \sqrt{3}\) and verify the relation between the zeroes and coefficients.
Solution:

Given \(\sqrt{3} x^2-8 x+4 \sqrt{3}\)

Let p(x) = \(\sqrt{3} x^2-8 x+4 \sqrt{3}=\sqrt{3} x^2-6 x-2 x+4 \sqrt{3}\)

\(\sqrt{3} x(x-2 \sqrt{3})-2(x-2 \sqrt{3})=(x-2 \sqrt{3})(\sqrt{3} x-2)\)

∴ p(x) = 0

⇒ \((x-2 \sqrt{3})(\sqrt{3} x-2)\) = 0

⇒ \(x-2 \sqrt{3}=0\)

or \(\sqrt{3} x-2=0\)

⇒ \(x=2 \sqrt{3}\)

or \(x=\frac{2}{\sqrt{3}}\)

∴ Zeroes of p(x) are \(2 \sqrt{3} \text { and } \frac{2}{\sqrt{3}}\)

Now, sum of zeros = \(=2 \sqrt{3}+\frac{2}{\sqrt{3}}\)

⇒ \(\frac{6+2}{\sqrt{3}}\)

⇒ \(\frac{8}{\sqrt{3}}\)

⇒ \(-\frac{(-8)}{\sqrt{3}}\)

⇒ \(-\frac{\text { coefficient of } x}{\text { coefficient of } x^2}\)

and product of zeros = \((2 \sqrt{3})\left(\frac{2}{\sqrt{3}}\right)=4\)

⇒ \(\frac{4 \sqrt{3}}{\sqrt{3}}\)

⇒ \(\frac{\text { constant term }}{\text { coefficient of } x^2}\)

Question 3. Find zeroes of the polynomial x2 – 4 and verify the relation between /crocs and coefficients.
Solution:

Given x2 – 4

Let p(x) = x2 – 4 = x2 – 22 = (x – 2) (x + 2)

∴ p(x) = 0

(x-2)(x + 2) = 0

x – 2 = 0 or x + 2 = 0

x = 2 or x = -2

∴ Zeroes of p(x) are 2 and -2

Now, sum of zeroes = 2 + (-2) = 0 = \(-\frac{0}{1}=-\frac{\text { coefficient of } x}{\text { coefficient of } x^2}\)

and product of zeroes = (2) (-2) = -4 = \(\frac{-4}{1}=\frac{\text { constant term }}{\text { coefficient of } x^2}\)

Question 4. Find a quadratic polynomial, the sum of whose zeroes is 5 and their product is 6. Hence, find the zeroes of the polynomial.
Solution:

Given

The sum of whose zeroes is 5 and their product is 6

Let a and (3 be the zeroes of the polynomial p(x).

Given that α + β = 5 and αβ = 6

Now, p(x) – x2 – (α + β)x + αβ = x2 – 5x ÷ 6 = x2 – 3x – 2x ÷ 6

= x(x – 3) – 2 (x – 3) = (x – 3) (x – 2)

There may be so many different polynomials which satisfy the given condition. Actually the general quadratic polynomial will be k(x2 – 5x + 6), where k = 0

∴ p(x) = 0

⇒ (x – 2)(x – 3) = 0

⇒ (x – 2) = 0 or (x – 3) = 0

⇒ x = 2 or x = 3

Zeroes are 2 and 3

zeroes of the polynomial 2 and 3.

Question 5. If the product of zeroes of the polynomial (ax2 – 6x – 6) is 4. find the value of a.
Solution:

Given polynomial = ax2 – 6x – 6

product of zeroes = \(\frac{\text { constant term }}{\text { coefficient of } x^2}\)

⇒ \(4=\frac{-6}{a}\)

4a = -6

⇒ \(a=-\frac{6}{4}\)

⇒ \(=-\frac{3}{2}\)

The value of a =\( -\frac{3}{2}\)

Question 6. If x = \(\frac{2}{3}\) and .v = -3 are zeroes of the quadratic polynomial ax2+7x+b, then find the values of a and b.
Solution:

Given

x = \(\frac{2}{3}\) and .v = -3 are zeroes of the quadratic polynomial ax2+7x+b

Let p(x) = ax2 + 7x + b

∵ \(x=\frac{2}{3}\) and x = -3 are zeroes of p(x)

⇒ \(p\left(\frac{2}{3}\right)=0\)

⇒ \(a\left(\frac{2}{3}\right)^2+7\left(\frac{2}{3}\right)+b=0\)

⇒ \(\frac{4 a}{9}+\frac{14}{3}+b=0\)

⇒ \(b=-\frac{4 a}{9}-\frac{14}{3}\)

and p(-3) = 0

⇒ a(-3)2+7(-3)+b = 0

⇒ \( 9 a-21-\frac{4 a}{9}-\frac{14}{3}=0\)

⇒ \(\frac{81 a-4 a}{9}=\frac{14}{3}+21=\frac{14+63}{3}\)

⇒ \(\frac{77 a}{9}=\frac{77}{3}\)

a=3

For equation (1)

⇒ \(b=\frac{-4 \times 3}{9}-\frac{-14}{3}=-6\)

a=3, b=-6

The values of a and b are 3 and -6.

Question 7. If one zero of the polynomial (a2 + 9)x2 + 13x+6a is reciprocal of the other, find the value of a.
Solution:

Given

one zero of the polynomial (a2 + 9)x2 + 13x+6a is reciprocal of the other

Let p(x) = (a2 + 9)x2 +13x+ 6a

Let α and \(\frac{1}{\alpha}\) be zeroes of p(x) then,

product of zeroes = \(\frac{\text { constant term }}{\text { coefficient of } x^2}\)

⇒ \(\alpha \cdot \frac{1}{\alpha}=\frac{6 a}{a^2+9}\)

⇒ \(I=\frac{6 a}{a^2+9}\)

⇒ a2+9 = 6a

⇒ a2-6a+9 = 0

⇒ (a-3)2 = 0

⇒ a-3 = 0

a = 3

The value of a is 3.

Relation Between Zeroes And Coefficients Of A Cubic Polynomial

Consider a cubic polynomial.

p(x) = ax3 + bx2 + cx + d, a ≠ 0

Let α, β, γ be zeroes of p(x), then (x – α), (x – β), (x – γ) will be the factors of p(x).

∴ ax3 + bx2 + cx + d = k(x – α)(x- β)(x – γ)

Comparing, we get

k = a

⇒ \(-k(\alpha+\beta+\gamma)=b \quad \Rightarrow \quad \alpha+\beta+\gamma=-\frac{b}{a}\)

(k=a)

⇒ \(k(\alpha \beta+\beta \gamma+\gamma \alpha)=c \quad \Rightarrow \quad \alpha \beta+\beta \gamma+\gamma \alpha=\frac{c}{a}\)

⇒ \(-k \alpha \beta \gamma=d \quad \Rightarrow \quad \alpha \beta \gamma=-\frac{d}{a}\)

p(x) = \(x^3-(\alpha+\beta+\gamma) x^2+(\alpha \beta+\beta \gamma+\gamma \alpha) x-\alpha \beta \gamma .\)

Division Algorithm For Polynomials

If(x) and g(x) are two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x), such that

f(x) = q{x) x g(x) + r(x).

where r(x) = 0 or deg {r(A)} < deg {g(x)}.

Division Algorithm Solved Examples

Question 1. Verify that 1,2,3 is the zeros of the cubic polynomial p(x) = x3-6x2+11x-6 and verify the relation between its zeroes and coefficients.
Solution:

Given p(x) = x3-6x2+11x-6

Here,

p(x)=\(x^3-6 x^2+11 x-6\)

p(1)=\(1^3-6(1)^2+11(1)-6=1-6+11-6=0\)

p(2)=\(=2^3-6(2)^2+11(2)-6=8-24+22-6=0\)

p(3)=\(=3^3-6(3)^2+11(3)-6=27-54+33-6=0\)

and

∴ 1,2 and 3 are zeroes of p(x).

Now, \(\alpha+\beta+\gamma\)

⇒ \(1+2+3=6\)

⇒ \(-\frac{-6}{1}\)

⇒ \(-\frac{\text { coefficient of } x^2}{\text { coefficient of } x^3}\)

αβ + βγ+ γα = (1 )(2) + (2)(3) + (3)(1) = 2 + 6 + 3= 11

⇒ \(\frac{11}{1}\)

⇒ \(\frac{\text { coefficient of } x}{\text { coefficient of } x^3}\)

⇒ \(\text { and } \alpha \beta y=(1)(2)(3)=6=-\frac{-6}{1}=-\frac{\text { constant term }}{\text { coefficient of } x^3}\)

Question 2. Find a cubic polynomial whose zeroes are \(\frac{1}{2},-\frac{3}{2}\) and 2
Solution:

Given

a cubic polynomial whose zeroes are \(\frac{1}{2},-\frac{3}{2}\) and 2

Let \(\alpha=\frac{1}{2}, \beta=-\frac{3}{2} \text { and } \gamma=2\)

∴ \(\alpha+\beta+\gamma=\frac{1}{2}-\frac{3}{2}+2=1\)

⇒ \(\alpha \beta+\beta \gamma+\gamma \alpha=\left(\frac{1}{2}\right)\left(-\frac{3}{2}\right)+\left(-\frac{3}{2}\right)(2)+\left(\frac{1}{2}\right)(2)=\frac{-3}{4}-3+1=-\frac{11}{4}\)

⇒ \(\alpha \beta \gamma=\left(\frac{1}{2}\right)\left(-\frac{3}{2}\right)(2)=-\frac{3}{2}\)

Cubic polynomial = \(x^3-(\alpha+\beta+\gamma) x^2+(\alpha \beta+\beta \gamma+\gamma \alpha) x-\alpha \beta \gamma\)

⇒ \(x^3-x^2-\frac{11}{4} x-\frac{3}{2} \approx 4 x^3-4 x^2-11 x-6\)

The cubic polynomial is \(x^3-x^2-\frac{11}{4} x-\frac{3}{2} \approx 4 x^3-4 x^2-11 x-6\)

Question 3. Divide 2x2+x-5 by x + 2 and verify the division algorithm.
Solution:

Given  2x2+x-5 and x + 2

Polynomials The Division Algorithm

Now, quotient = 2x- 3, remainder = 1

dividend = 2x2 + x- 5 and divisor = x+ 2

and quotient x division + remainder = (2x-3)(x+2) +1

\(2 x^2+4 x-3 x-6+1=2 x^2+x-5\)

= division

Question 4. If the polynomial \(\left(x^4+2 x^3+8 x^2+12 x+18\right)\) is divided by another polynomial (x2+5), the remainder comes out to be (px+q). Find the values of p and q.
Solution:

Given \(\left(x^4+2 x^3+8 x^2+12 x+18\right)\) is divided by another polynomial (x2+5)

Let f(x) = \(x^4+2 x^3+8 x^2+12 x+18\) and g(x) = x2+5

On dividing f(x) by g(x)

Polynomials Find The Values Of P And Q

∵ the remainder is given px+q.

∴ p=2 and q=3

The values of p and q 2 and 3.

Question 5. What real number should be subtracted from the polynomial (3x3 + 10x2 – 14x + 9) so that (3x- 2) divides it exactly?
Solution:

Given (3x3 + 10x2 – 14x + 9)

On dividing (3x3 + 10x2 – 14x + 9) by (3x- 2)

Polynomials Real Number Should Be Subtracted From The Polynomial

∵ the remainder = 5

∴ required number = 5

The real number should be subtracted from the polynomial is 5.

Question 6. If 2 is a zero of the polynomial x3-2x2-x+2, then find its other zeroes.
Solution:

Given

2 is a zero of the polynomial x3-2x2-x+2

Let p(x) = x3-3x3-x+2

∵ x=2 is a zero of p(x)

(x-2) is a factor of p(x)

Polynomials If 2 Is A Zero Of The Polynomial

∴ p(x) = \(x^3-2 x^2-x \div 2=(x-2)\left(x^2-1\right)=(x-2)\left(x^2-1^2\right)\)

= (x-2)(x-1)(x 1)

Now, p(x) = 0

⇒ (x-2)(x-1)(x 1) = 0

⇒ x-2 = 0 or x-1 = 0  or x  1 =0

⇒ x=2 or x=1 or x=-1

Hence, other zeroes are 1 and -1.

Question 7. Obtain all other zeroes of (x4 + 4x3– 2x2– 20x- 15) if two of its zeroes are \(\sqrt{5} \text { and }-\sqrt{5}\)
Solution:

Given

(x4 + 4x3– 2x2– 20x- 15) if two of its zeroes are \(\sqrt{5} \text { and }-\sqrt{5}\)

Let p(x) = x4 + 4x3– 2x4– 20x- 15

\(\sqrt{5} \text { and }-\sqrt{5}\) are zeroes of p(x)

∴ \(x-\sqrt{5} \text { and } x+\sqrt{5}\) are factors of p(x).

So, \((x-\sqrt{5})(x+\sqrt{5})=x^2-5\) is a factor of p(x)

Polynomials If Two Of Its Zeroes Are Root 5 And Minus Root 5

∴ p(x) = \(x^4+4 x^3-2 x^2-20 x-15=\left(x^2-5\right)\left(x^2+4 x+3\right)\)

⇒ \(\left(x^2-5\right)\left[x^2+x+3 x+3\right]=\left(x^2-5\right)[x(x+1)+3(x+1)]\)

⇒ \(\left(x^2-5\right)(x+1)(x+3)\)

∴ The other zeroes are given by

x+1 = 0 or x+3 = 0

⇒ x = -1 or x = -3

Hence, other zeroes are -1 and -3

Question 8. Find zeroes of the polynomial p(x) = x3 – 9x2 + 26x – 24, if it is given that the product of its two zeroes is 8.
Solution:

Given polynomial p(x) = x3 – 9x2 + 26x – 24

Let α, β, γ be zeroes of the given polynomial p(x), such that αβ = 8 …….(1)

⇒ \(\alpha+\beta+\gamma=-\frac{(-9)}{1}=9\) ……….(2)

⇒ \(\alpha \beta+\beta \gamma+\gamma \alpha=\frac{26}{1}=26\) …..(3)

⇒ \(\alpha \beta \gamma=-\frac{(-24)}{1}=24\) ………..(4)

From (1) and (4)

8γ = 24

γ = 3

Put γ = 3 in (2), we get α+β = 6 ……..(5)

⇒ \((\alpha-\beta)^2=(\alpha+\beta)^2-4 \alpha \beta\)

⇒ \(=(6)^2-4(8)\)

⇒ \(=36-32\)

⇒ \(=4\)

⇒ \(\alpha-\beta= \pm 2\) …..(6)

Solving (5) and (6), we get

a= 4, b = 2  or a=2, b = 4 and r = 3

So, zeroes are 2,3 and 4

Question 9. Find the common zeroes of the polynomials x3 +x2– 2x- 2 and x3– x2 – 2x + 2.
Solution:

Given x3 +x2– 2x- 2 and x3– x2 – 2x + 2

For the common zeroes, first, we find the H.C.F. of given polynomials by Euclid’s division method (long division method)

Polynomials First We Find The HCF Of Given Polynomials By Euclids Division Method

Hence, the H.C.F. of the given polynomials is (x2 – 2). Thus, the common zeroes of the given polynomials are the zeroes of x2– 2 i.e. \((x+\sqrt{2})(x-\sqrt{2})\)

So, zeroes of \(\sqrt{2} \text { and }-\sqrt{2}\).

Polynomials Exercise 2.1

Question 1. The graphs o(y = p(x) are given below, for some polynomials p{x). Find the number of zeroes of p(x), in each case.

Polynomials Find The Number Of Zeroes Of P Of X

Polynomials Find The Number Of Zeroes Of P Of X.

Solution:

1. The graph of the polynomial p(x) does not intersect the x-axis at any point.

∴ Number of its zeroes is zero.

2. The graph of the polynomial p(x) intersects the x-axis at one point.

∴ Number of its zeroes is one.

3. The graph of the polynomial p(x) intersects the x-axis at three points.

∴ Number of its zeroes is 3.

4. The graph of the polynomial p(x) intersects the x-axis at two points.

∴ Number of its zeroes is 2

5. The graph of the polynomial p(x) intersects the x-axis at four points.

∴ Number of its zeroes is 4.

6. The graph of the polynomial p(x) intersects the x-axis at three points.

∴ Number of its zeroes is 3.

Polynomials Exercise 2.2

Question 1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:

  1. x2 – 2x- 8
  2. 4s2 – 4s + 1
  3. 6x2 – 3 – 7x
  4. 4u2 + 8u
  5. t2– 15
  6. 3x2 -x- 4

Solution:

Let p(x) =x2 – 2x- 8

= x2-4x + 2x-8

= x(x- 4) + 2(x- 4)

= (x- 4)(x + 2)

The zeroes of p(x) will be given by x- 4 = 0 and x + 2 = 0.

x- 4 = 0 ⇒ x = 4

x + 2 = 0 ⇒ x = -2

∴ Zeroes of p(x) = 4,-2

Now, from p(x) = x2 – 2x- 8

a = 1 , b = -2, c = -8

∴ \(-\frac{b}{a}=-\frac{(-2)}{1}=2\)

= 4+(-2) = sum of zeroes

and \(\frac{c}{a}=\frac{-8}{1}=-8\)

= 4(-2) = product of zeroes

∴ Relations between the zeroes of polynomial and the coefficients are true.

2. Let p(s) = 4s2 – 4s +1 = 4s2-2s-2s +1

= 2s(2s-1) -1 (2s-1)

= (2s-1)(2s-1)

The zeroes ofp(s) will be given by 2s -1 = 0 and 2s – 1 = 0.

∴ 2s – 1 = 0 \(s=\frac{1}{2}\)

Therefore, zeroes of p(s) = \(\frac{1}{2}, \frac{1}{2}\)

Now, from p(s) = 4s2 – 4s + 1

a = 4, b =-4, c = 1

⇒ \(-\frac{b}{a}=-\frac{(-4)}{4}=1\)

⇒ \(\frac{1}{2}+\frac{1}{2}\)

=  sum of zeroes

and \(\frac{c}{a}=\frac{1}{4}=\frac{1}{2} \times \frac{1}{2}\) = Product zeroes.

Relations between the zeroes of polynomial and the coefficients are true.

3. Let p(x) = 6x2 – 3 -7x = 6x2 – 7x- 3

= 6x2– 9x + 2x- 3

= 3x(2x-3) + 1(2x- 3)

= (2r- 3)(3x+ 1)

The zeroes of p(x) will be given by 2x- 3 = 0 and 3x+ 1=0.

∴ 2x-3 = 0 ⇒ \(x=\frac{3}{2}\)

and 3x + 1 = 0 ⇒ \(x=-\frac{1}{3}\)

Therefore, zeroes of p(x) = \(\frac{3}{2},-\frac{1}{3}\)

Now, from p(x) =  6×2- 7x- 3

a = 6, b=-7, c = -3

⇒ \(-\frac{b}{a}=-\frac{(-7)}{6}=\frac{7}{6}=\frac{3}{2}+\left(-\frac{1}{3}\right)\)

= sum of zeroes.

⇒ \(\frac{c}{a}=\frac{-3}{6}=-\frac{1}{2}=\frac{3}{2} \times\left(-\frac{1}{3}\right)\)

= product of zeroes.

Therefore, the relations between the zeroes of the polynomials and the coefficients are true.

4. Let p(u)= 4u2 + 8u = 4u(u + 2) Zeroes of p(u) will be given by 4u = 0 and u + 2 = 0.

∴ 4u = 0 ⇒ u = 0

and u + 2 = 0  ⇒ u = -2

Therefore, zeroes of p(u) = 0, -2.

Now, from p(u) = 4u2 + 8u, a = 4, b = 8, c = 0

∴ \(-\frac{b}{a}=-\frac{8}{4}=-2=0+(-2)=\text { sum of zeroes. }\)

and \(\frac{c}{a}=\frac{0}{4}=0=0(-2)=\text { product of zeroes. }\)

Therefore, the relations between the zeroes of the polynomials and the coefficients are true.

5.  Let p(t) = t2 – 15

⇒ \(t^2-(\sqrt{15})^2=(t-\sqrt{15})(t+\sqrt{15})\)

The zeroes of p{t) will be given by

⇒ \(t-\sqrt{15}=0\) and \(t-\sqrt{15}=0\)

Now, \(t-\sqrt{15}=0\) \(\Rightarrow \quad t=\sqrt{15}\)

and \(t-\sqrt{15}=0\) \(\Rightarrow t=-\sqrt{15}\)

Zeroes of p(t) = \(\sqrt{15} \text { and }-\sqrt{15}\)

From p(t) = t2 – 15

a = 1, b = 0, c =-l5

⇒ \(-\frac{b}{a}=-\frac{0}{1}=0=\sqrt{15}+(-\sqrt{15})\)

= sum of zeroes.

and \(\frac{c}{a}=\frac{-15}{1}=-15=(\sqrt{15})(-\sqrt{15})\)

= product of zeroes.

Therefore, the relations between the zeroes of the polynomials and the coefficients are true.

6.  Let p(x)  = \(3 x^2-x-4=3 x^2-4 x+3 x-4\)

⇒ \(x(3 x-4)+1(3 x-4)\)

⇒ \((3 x-4)(x+1)\)

The zeroes of p(x) will be given by 3x – 4 = 0 and x + 1 = 0.

Now, 3x- 4 = 0 ⇒ \(x=\frac{4}{3}\)

and x+ 1 = 0 ⇒ A =-l

Zeroes of p(x) = \(\frac{4}{3}\) and -1

From p(x) = 3x2-x-4

a = 3, b = -1, c = -4

⇒ \(-\frac{b}{a}=-\frac{(-1)}{3}=\frac{1}{3}=\frac{4}{3}+(-1)\)

= sum of zeroes.

and \(\frac{c}{a}=\frac{-4}{3}=\frac{4}{3} \times(-1)=\text { sum of product. }\)

Therefore, the relations between the zeroes of the polynomials and the coefficients are true.

Question 2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively:

  1. \(\frac{1}{4},-1\)
  2. \(\sqrt{2}, \frac{1}{3}\)
  3. \(0, \sqrt{5}\)
  4. 1,1
  5. \(-\frac{1}{4}, \frac{1}{4}\)
  6. 4,1

Solution:

1. Let the zeroes of the polynomial be α and β.

∴ \(\alpha+\beta=\frac{1}{4} \text { and } \alpha \beta=-1\)

∴ quadratic polynomial = (x- α) (x- β)

= x2 – (α+ β)x + αβ

= \(x^2-\frac{1}{4} x-1=\frac{1}{4}\left(4 x^2-x-4\right)\)

∴ Required polynomial = 4x2 – x- 4.

2. Let the zeroes of the polynomial be a and p.

∴ \(\alpha+\beta=\sqrt{2} \quad \text { and } \quad \alpha \beta=\frac{1}{3}\)

∴ quadratic polynomial = (x – α)(x – β)

⇒ \(x^2-(\alpha+\beta) x+\alpha \beta=x^2-\sqrt{2} x+\frac{1}{3}\)

⇒ \(\frac{1}{3}\left(3 x^2-3 \sqrt{2} x+1\right)\)

∴ Required polynomial = \(3 x^2-3 \sqrt{2} x+1\)

3. Let the zeroes of the polynormal be a and p

∴ \(\alpha+\beta=0 \text { and } \alpha \beta=\sqrt{5}\)

∴ quadratic polynomial = (x – α)(x – β)

⇒ \(x^2-(\alpha+\beta) x+\alpha \beta\)

⇒ \(x^2-0(x)+\sqrt{5}=x^2+\sqrt{5}\)

4.  Let the zeroes of the polynomial be a and p.

α + β = 1 and αβ = 1

∴ quadratic polynomial = (x- a)(x- P)

= x2 – (α + β)x + αβ

x2– x + 1

5.  Let the zeroes of the polynomial be a and p.

∴ \(\alpha+\beta=-\frac{1}{4} \text { and } \alpha \beta=\frac{1}{4}\)

Now, quadratic polynomial = (x – α)(x – β)

x2 – (α + β)x + αβ

⇒ \(\alpha+\beta=-\frac{1}{4} \text { and } \alpha \beta=\frac{1}{4}\)

Therefore, required polynomial = 4x2 + x + 1

6. Let the zeroes of the polynomial be α and β.

∴ α + β = 4 and αβ = 1

Now, quadratic polynomial = (x- α)(x- β)

= x2 — (α + β)x + αβ

= x2-4x+ 1.

Polynomials Exercise 2.3

Question 1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following :

  1. p(x) =x3– 3x2 + 5x-3, g(x) =x2-2
  2. p(x) = x4 – 3x2 + 4x + 5, g(x) =x2+1 -x
  3. p(x) = x4 – 5A + 6, g(x) = 2 -x2

Solution:

Polynomials Divide The Polynomial P Of X And G Of X

Now, P(x) = g(x). q(x) + r(x)

∴ quotient q(x) = x- 3

and remainder r(x) = 7x- 9

Polynomials Divide The Polynomial P Of X By The Polynomial Of G Of X 2

Now, P(x) = g(x) f(x) + r(x)

∴ quotient q(x) = x2 +x-3

and remainder r(x) = 8

Polynomials Divide The Polynomial P Of X By The Polynomial Of G Of X 3

Now, p(x) = g(x) q(x) + r(x)

∴ quotient q(x) = -x2 – 2

and remainder r(x)= -5x + 10

Question 2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

  1. t2-3, 2t4 + 3t3– 2t2– 9t- 12
  2. x2 + 3x + 1, 3x4 + 5x3– 7x2 + 2x + 2
  3. x3– 3x + 1, x5 – 4x3 + x2 + 3x +1

Solution:

Polynomials Dividing The First And Second Polynomial 1

Now, p(t) g(t) + (t) + r(t)

∴ quotient q(t) = 2t2 + 3t + 4 and remainder r(x)= 0.

∵ The remainder is zero.

∴ t2 – 3 is a factor of 2t4 + 3t3 -2t2 – 9t – 12.

Polynomials Dividing The First And Second Polynomial 2

Now, p= g(x) q(x) + r(x)

∴ quotient q(x) = 3x2 – 4x + 2 and remainder r(x) = 0

∵ The remainder is zero.

∴ x2 + 3x + 1 is a factor of 3x4 + 5x3– 7x2 + 2x + 2.

Polynomials Dividing The First And Second Polynomial 3

Now, P(x) = g(x) q(x) + r(A)

∴ quotient q(x) = x2– 1 and remainder r(x) = 2.

∵ The remainder is not zero.

∴ x2– 3x + 1 is not a factor x5 – 4x3 + x2 + 3x + 1 .

Question 3. Obtain all other zeroes of 3×4 + 6×3 – 2×2 – 10x – 5 if two of its zeroes are \(\sqrt{\frac{5}{3}} \text { and }-\sqrt{\frac{5}{3}}\)
Solution:

Given 3x4 + 6x3 – 2x2 – 10x – 5

Two of its zeroes are \(\sqrt{\frac{5}{3}} \text { and }-\sqrt{\frac{5}{3}}\)

Let the remaining two zeroes be α and β.

Given zeroes are \(\sqrt{\frac{5}{3}} \text { and }-\sqrt{\frac{5}{3}}\)

Quadratic polynomial from zeroes.

⇒ \(\left(x-\sqrt{\frac{5}{3}}\right)\left(x+\sqrt{\frac{5}{3}}\right)\)

⇒ \(x^2-\frac{5}{3}=\frac{1}{3}\left(3 x^2-5\right)\)

∴ quadratic polynomial = 3x2 – 5

Now, (x- α)(x- β)(3x2 – 5)

⇒ \(3 x^4+6 x^3-2 x^2-10 x-5\)

⇒ \((x-\alpha)(x-\beta)=\frac{3 x^4+6 x^3-2 x^2-10 x-5}{3 x^2-5}\)

Polynomials Quadratic Polynomial

∴ (x- α)(x- β) = x2 + 2x + 1

= (x+ 1)(x + 1)

∴ α = -1, β =-1

⇒ Remaining zeroes =-1,-1.

Question 4. On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x- 2 and -2x + 4, respectively. Find g(x).
Solution:

Given

x3 – 3x2 + x + 2

Let p(x) = x3 – 3x2 + x + 2

quotient q(x) =x – 2 and remainder r(x)= -2x + 4

Now, p(x) = g(x) . q(x) + r(x)

⇒ g(x)-q(x) = p(x)- r(x)

= (x3 – 3x2 + x + 2) – (-2x + 4)

= x3 – 3x2 + x + 2 + 2x- 4

= x3 – 3x2 + 3x- 2

⇒ g(x) = \(\frac{x^3-3 x^2+3 x-2}{x-2}\) (∵ q (x) = x-2)

Polynomials On Dividing Polynomial G Of X

Therefore, g(x) = x2-x+1

Question 5. Give examples of polynomials p(x), g(A), q(x) and r(x), which satisfy the division algorithm and

  1. Deg P(x) = deg q(x)
  2. Deg q(x) = deg r(x)
  3. Deg r(x) = 0

Solution:

1. deg p(x) = deg q(x)

We know that

p(x) = g(x) q(x) + r(x)

∴ degree of g(x) = zero

Let p(x) = 2x3 + 6x2 + 2x- 1

and g(x) = 2

Polynomials Give Examples Of Polynomials 1

∴ q(x) = x3 + 3x2 + x and r(x) = -1

So, p(x) = 2x3 + 6x2 + 2x- 1, = 2,

q(x) = x3 + 3x2 + x, r(x) = -1

degree q(x) = degree r(x)

We know that

p(x) = q(x) +r(x)

If q(x) and r(x) are polynomials of the first degree then a degree of p(x) must be 1 more than the degree: of g(x)

Let p(x) =  x3 + 3x2 + 2x + 5 and g(x) = x2 – 1

Polynomials Give Examples Of Polynomials 2

∴ q(x) = x + 3 and r(x) = 6x + 5

so, p(x) = x3 + 3x2 + 2x + 5, g(x) =x2-1

q(x) = x + 3, r(x) = 6x + 5

3. degree r(x) = 0

For this, the degree of g(x) must be 1.

Let p(x) = 2x3 – 3×2 +x+ 4 and g(x) = x- 1 .

Polynomials Give Examples Of Polynomials 3

∴ q(x) = 2x2 – x and r(x) = 4

Therefore, p(x) = 2x3 – 3x2 + x + 4, g(x) = x- 1,

q(x) = 2x2 – x, r(x) = 4

Polynomials Exercise 2.4 (Optional)

Question 1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case :

  1. \(2 x^3+x^2-5 x+2 ; \frac{1}{2}, 1,-2\)
  2. \(x^3-4 x^2+5 x-2 ; 2,1,1\)

Solution:

1. Let p(x) = \(2 x^3+x^2-5 x+2\)

∴ \(p\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^2-5\left(\frac{1}{2}\right)+2\)

⇒ \(\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2=0\)

⇒ \(\frac{1}{2}\)is a zero of p(x).

Again ,p(1) = 2(1)3 + (1)2 – 5(1) + 2

=2 + 1 – 5 + 2 = 0

⇒ 1 is a zero of p(A).

Again p(-2) = 2(-2)3 + (-2)2 – 5(-2) + 2

= -16 + 4+10 + 2 = 0

⇒ -2 is a zero of p(x)

Now, in p(x) = 2x3 +x2 -5x + 2

a = 2,b = 1 ,c = -5, d = 2

⇒ \(-\frac{b}{a}=-\frac{1}{2}=\frac{1}{2}+1+(-2)=\text { sum of zeroes }\)

⇒ \(\frac{c}{a}=\frac{-5}{2}=\frac{1}{2} \times 1+1 \times(-2)+\frac{1}{2} \times(-2)\)

= sum of the product of zeroes taken two at a time.

and \(-\frac{d}{a}=\frac{-2}{2}=-1\)

⇒ \(\frac{1}{2} \times 1 \times(-2)=\text { product of zeroes}\)

Therefore the relations between the zeroes of the polynomial and the coefficients are true.

2. Let p(x) = x3 – 4x2 + 5x -2

p(2) = (2)3 – 4(2)2 + 5(2)-2

= 8-16+10-2 = 0

⇒ 2 is a zero of p(x).

Again P(1) = (1)3 – 4(1)2 + 5(1)-2

= 1-4 + 5- 2 = 0

= 1 is a zero of p(x).

Now, for p(x) = x3 – 4x2 + 5x- 2

a= 1, b =-4, c = 5, d = -2

∴ \(-\frac{b}{a}=-\frac{(-4)}{1}=4=2+1+1=\text { sum of zeroes. }\)

⇒ \(\frac{c}{a}=\frac{5}{1}=5=2 \times 1+1 \times 1+1 \times 2\)

= sum of the product of zeroes taken two at a time.

and \(-\frac{d}{a}=-\frac{(-2)}{1}=2=2 \times 1 \times 1\)

= product of zeroes.

Therefore the relations between the zeroes of the polynomial and the coefficients are true.

Question 2. Find a cubic polynomial with the sum, the sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, and -14 respectively.
Solution:

Given

The sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, and -14 respectively

Let the zeroes be α, β, and γ.

∴ α + β + γ = 2

αβ + βγ + γα = -7 and αβγ = -14

∴ Cubic polynomial = (x-α)(x-β)(x-γ)

= x3 – (α+ β + γ)x2+ (αβ + βγ + γα)x- αβPγ

= x3-2x2-7x+ 14

Question 3. If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, find a and b.
Solution:

Given

x3 – 3x2 + x + 1

a – b, a, a + b, are zeroes of polynomial.

Let p(x) = x2 – 3x2 +x + 1

⇒ A= 1, B =-3, C = 1,D = 1

sum of zeroes = \(-\frac{B}{A}\)

a-b + a + a + b = \(-\frac{(-3)}{1}\)

3a = 3

a= 1

and product of zeroes = \(-\frac{D}{A}\)

⇒ \((a-b) \cdot a \cdot(a+b)=-\frac{1}{1}\)

⇒ \(a\left(a^2-b^2\right)=-1\)

⇒ \(1-b^2=-1\)

⇒ \(b^2=2\)

⇒ \(b= \pm \sqrt{2}\)

∴ \(a=1, b= \pm \sqrt{2}\)

Question 4. If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x- 35 are \(2 \pm \sqrt{3}\) , find other zeroes.
Solution:

Given

x4 – 6x3 – 26x2 + 138x- 35

Two zeroes of a polynomial are \(2+\sqrt{3} \text { and } 2-\sqrt{3} \text {. }\)

Let the remaining two zeroes be a and p

∴ \((x-\alpha)(x-\beta)(x-2-\sqrt{3})(x-2+\sqrt{3})\)

⇒ \(x^4-6 x^3-26 x^2+138 x-35\)

⇒ \(\quad(x-\alpha)(x-\beta)\left[(x-2)^2-(\sqrt{3})^2\right]\)

⇒ \(x^4-6 x^3-26 x^2+138 x-35\)

⇒ \(\quad(x-\alpha)(x-\beta)\left(x^2-4 x+4-3\right)\)

⇒ \(x^4-6 x^3-26 x^2+138 x-35\)

⇒ \(\quad(x-\alpha)(x-\beta)\)

⇒ \(\quad\frac{x^4-6 x^3-26 x^2+138 x-35}{x^2-4 x+1}\)

Polynomials If Two Zeroes Of The Polynomial

∴ (x -α) (x- β) = x2 – 2x- 35 = x2– 7x + 5x- 35

= x(x- 7) + 5(x- 7)

= (x- 7)(x + 5)

The other zeroes are α = 7, β = -5

Question 5. If the polynomial x4 – 6x3 + 16x2-25x+10 is divided by another polynomial x2-2x + k. the remainder comes out to be x + a, find k and a.
Solution:

Given

x4 – 6x3 + 16x2-25x+10 and x2-2x + k

Let p(x) = x4 – 6x3 + 16x2 – 25x + 10 divisor g(x) = x2 -2x + k and remainder r(x) = x + a

Polynomials If The Polynomial Is Divided By Another Polynomial

According to the problem,

(2k – 9)x +(10-8k + k2) = x + a Comparing the coefficient of x

2k-9= 1

⇒ 2k = 1 + 9= 10

⇒ k = 5

Comparing the constant terms

10 – Sk + k2 = a

⇒ 10-40 + 25 = a  (put K = 5)

⇒ a = -5

So, k = 5 and a =-5

Polynomials Multiple Choice Questions

Question 1. If one zero of the polynomial 3x2 + x- k is 3 then the value of k is :

  1. -30
  2. -24
  3. 30
  4. 24

Answer: 3. 30

Question 2. A polynomial with zeroes 2 and -3 is:

  1. x2 -x- 6
  2. x2 + x- 6
  3. x2 – 6
  4. x2 + 6

Answer: 2. x2 + x- 6

Question 3. The number of polynomials with zeroes 4 and 3, is :

  1. 1
  2. 2
  3. 3
  4. Infinite

Answer: 4. Infinite

Question 4. The zeroes of x2 + 6a + 5 are:

  1. Both positive
  2. Both negative
  3. Both equal
  4. One zero is zero

Answer: 2. Both negative

Question 5. If two zeroes of the polynomial ax3 + bx2 + cx + d are zero, then third zero is :

  1. \(-\frac{b}{a}\)
  2. \(\frac{b}{a}\)
  3. \(\frac{c}{a}\)
  4. \(-\frac{c}{a}\)

Answer: 1. \(-\frac{b}{a}\)

Question 6. If x6– 1 is divided by a polynomial of third degree, the maximum degree of the remainder can be:

  1. 0
  2. 1
  3. 2
  4. 3

Answer: 3. 2

Question 7.  The product of zeros of ax2 + bx + c, a≠0 is:

  1. \(-\frac{b}{a}\)
  2. \(\frac{b}{a}\)
  3. \(-\frac{c}{a}\)
  4. \(\frac{c}{a}\)

Answer: 4. \(\frac{c}{a}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 1 Real Numbers

Real Numbers Introduction

We have studied real numbers. Now we will discuss two important results, namely

  1. Euclid’s division lemma
  2. Fundamental theorem of arithmetic

NCERT Exemplar Solutions For Class 10 Maths Chapter 1 Real Numbers

Difference Between Algorithm And Lemma

Algorithm: An algorithm is a series of well-defined steps, that gives a procedure for solving a problem.

Lemma: It is a proven statement used to prove another statement.

Euclid’s Division Lemma

Real Numbers Euclids Division Lemma

Note: Remainder is always less Dividend than the divisor. It is greater = Divisor x Quotient+ Remainder than or equal to zero.

i.e., \(0 \leq r<\text { divisor }\)

We think that all of you are familiar with this well-known procedure. Now, we will generalize this division method known as Euclid’s Divison Lemma.

Statement

For any two given positive integers a and b, there exists unique whole numbers q and r such that

a = bq +r

Real Numbers Positive Integers A and B There Exists Unique Whole Numbers Q and R

Where 0 ≤ r< b

Here, a = dividend, b = divisor, q = quotient, r= remainder

Euclid’s Division Algorithm (To Find The HCF Of Two Positive Integers)

It is very useful to obtain the H.C.F. of two positive integers. Let c and d be two positive integers with c > d.

Now, to find the H.C.F. of c and d, follow the following steps:

Step 1: Apply Euclid’s division lemma to c and d.

We find whole numbers q and r such that

c = dq + r

where, 0 ≤ r < d,

Step 2: If r = 0 then d (recent divisor) is the H.C.F. of c and d,  and if r ≠ 0 then, apply the division lemma to d and r.

Step 3: Continue the process till the remainder is zero. The divisor at this stage will be the required H.C.F.

Euclid’s Division Algorithm Solved Examples

Question 1. Use Euclid’s algorithm to find the H.C.F. of 4052 and 12570.
Solution:

Here, 12576 >4052

Real Numbers Euclids Division Algorithm

∴ 12576 = 4052 × 3 + 420

4052 = 420 × 9 + 272

420 =272 × 1 + 148

272 = 148 × 1 + 124

148 = 124 × 1 + 24

124 =24 × 5 + 4

24 = 4 × 6 + 0

Since remainder = 0, ⇒ recent divisor is the H.C.F.

∴ H.C.F. (12576, 4052)= 4

The H.C.F. of 4052 and 12570 = 4

Question 2. If the H.C.F. of 210 and 55 is expressible in the form 210×5 + 55x, then find the value of x.
Solution:

First, we will find the H.C.F. of 210 and 55.

Here, 210 >55

Real Numbers The HCF Of 210 And 55

210 = 55 × 3 + 45

55 =45 × 1 + 10

45 = 10 × 4 + 5

10 =5 × 2 + 0

Since remainder = 0

⇒ The recent divisor is the H.C.F.

∴ H.C.F. of (210. 55) = 5

Now, 210 × 5 + 55x = 5

55x = 5 – 210 x 5

= 5- 1050

=-1045

⇒ \(x=-\frac{1045}{55}\)

=-19

The value of x =-19

Question 3. Find die H.C.F. of 65 and 117 and express it in the form of 65x + 117y.
Solution:

Here, 117 > 65

Real Numbers The HCF Of 65 And 117

∴ 117 =65 × 1 + 52

65 = 52 × 1 + 13

52 = 13 × 4 + 0

Since remainder = 0

⇒ recent divisor is die H.C.F.

∴ H.C.F. (117, 65) = 13

Now, 13 = 65 -52 × 1

13 =65 – (117- 65 × 1)

= 65 – 117 + 65 × 1

= 65 × 2-117

= 65x + 117y

where, x = 2 and y = -1

Remark:

It follows from the above example that the H.C.F. (say h) of two positive integers o and b can be expressed as a linear combination of o and b i.e., h = xa + yb for some integers x and y. This representation is not unique. Because, h = xa + yb

⇒ \(h=x a+\underbrace{a b-a b}_{\begin{array}{c} \text { Add and } \\ \text { subtract } \end{array}}+y b=a(x+b)+b(y-a)\)

⇒ \(h=x a-\underbrace{a b+a b}_{\begin{array}{c} \text { Subtract } \\ \text { and add } \end{array}}+y b=a(x-b)+b(y+a)\)

So, the coefficients of a and b may be

  1. x and y
  2. x + band y-a
  3. x-b and y+ a

Hence, the linear combination of a and b is not unique.

Factor

It divides the number whose factor is this.

For example: 1,2,3 and 6 are the factors of 6. So, each factor divides 6 completely.

Multiple

It is divided by the number whose multiple is this.

For example : 3, 6, 9 12, 15 and 18 are the multiples of number 3. So, each multiple can be divided by the number 3.

Remember:

  1. A smaller number divides the larger while larger number is divided by smaller.
  2. Factors are smaller while multiples are larger.

H.C.F. (Highest Common Factor)

H.C.F. is the highest common factor of two or more numbers which divides each of the numbers.

For example: Two numbers are 24 and 36.

Real Numbers Highest Common Factor

Highest (maximum) common factor = 12

∴ H.C.F. = 12

So, 12 is the highest number which divides 24 and 36 completely.

L.C.M. (Least Common Multiple)

L.C.M. of two or more numbers is the least common multiple that is divided by all the numbers.

For example: Two numbers are 24 and 36.

Real Numbers Least Common Multiple

∴ The least (minimum) common multiple is 72.

So, 72 is the least number which is divided by 24 and 36 both.

Question 4. Find the largest number, which divides 246 and 1030 leaving the remainder 6 in each case.
Solution:

We have to find a number, which divides the other numbers means → H.C.F.

It is given that the required number, when divided between 246 and 1030 leaves the remainder 6 i.e.; 6 is extra in each number. It means that if these numbers are 6 less, then there is no remainder in each case.

∴ 246- 6 = 240 and 1030- 6 = 1024 are completely divisible by the required number.

Now, 1024 > 240

Real Numbers The Largest Number Divides 246 And 1030

∴ H.C.F. (1024,240)= 16

Hence, required no. = 16

Question 5. Find the largest number that will divide 400, 437, and 542 leaving the remainder 9, 12,15 respectively.
Solution:

We have to find a number, which divides the other numbers means → H.C.F.

It is given that the required number, when divided into 400, 437, and 542, leaves the remaining 9, 12, and 15 respectively. It means that if 400 is 9 less, 437 is 12 less, and 542 is 15 less, then on division, gives no remainder (extra).

∴ 400 – 9 = 391,437-12 = 425 and 542 – 15 = 527 are completely divisible by the required number.

First, we will find the H.C.F. of 391 and 425.

Real Numbers We Will Find The HCF Of 391 And 425

∴ 425 =391 × 1 +34

391 = 34 × 11 + 17

34 =17 × 2 + 0

∴ H.C.F. (391,425) = 17

Now, we will find the H.C.F. of 17 and 527.

Real Numbers We Will Find The HCF Of 17 And 527

527 = 17 × 31 +0

⇒ H.C.F. (17,527) = 17

∴ Required number = 17

Question 6. Show that one and only one out of n, (n+ 1) and (n + 2) is divisible by 3, where n is any positive integer.
Solution:

When n is divided by 3, let q and r be the quotient and remainder respectively.

∴ n = 3q + r

where, 0 ≤ r < 3 i.e.,

r = 0 or r = 1 or r = 2

1. When, r = 0, then

n = 3q,

which is divisible by 3.

n + 1 = 3q + 1

which leaves a remainder of 1 when divided by 3.

i. e., (n + 1) is not divisible by 3.

n + 2 =3q + 2

which leaves a remainder of 2 when divided by 3.

i.e., (n + 2) is not divisible by 3.

So, only n = 3q is divisible by 3 when r = 0.  …….(1)

2. When, r = 1, then

n= 3q + 1

which is not divisible by 3. (∵ it leaves a remainder 1)

n + 1 = 3q + 2

which is not divisible by 3. (∵ remainder = 2)

n + 2 = 3q + 3 = 3 (q + 1)

which is divisible by 3.

So, only n + 2 is divisible by 3 when r= 1.  ……….(2)

3. When, r = 2, then

n = 3q + 2

which is not divisible by 3. (∵ remainder = 2)

⇒ n + 1 = 3q + 3 = 3 (q + 1)

which is divisible by 3.

n + 2 =3q + 4 = 3q + 3 + 1= 3 (q + I) + 1

which is not divisible by 3 (∵ remainder = 1)

So, only n + 1 is divisible by 3 when r = 2.  ………(3)

Hence, from equations (1), (2) and (3), we can say that one and only one out of n, (n + 1) and (n + 2) is divisible by 3.

Hence Proved.

Question 7. There are 156, 208, and 260 students in groups A, B, and C respectively. Buses are to be hired to take them for a field trip. Find the minimum number of buses to be hired, if the same number of students should be accommodated in each bus, and if a separate bus for separate groups is needed.
Solution:

Given:

There are 156, 208, and 260 students in groups A, B, and C respectively. Buses are to be hired to take them for a field trip.

First of all we shall find the H.C.F. of number of students in each group. This will give tls the maximum number of students of the same group in each bus.

Given numbers are 156, 208 and 260. ,

Here, 260 > 208 > 156

Let us first find the H.C.F. of 156 and 208

By using Euclid’s division lemma for 156 and 208,

we get 208 = 156 × 1 + 52. Here remainder ≠ 0

156 = 52 × 3 + 0.

Here, remainder = 0

⇒ The recent divisor is the H.C.F.

∴ H.C.F. of 156 and 208 is 52.

Now, 260 > 52

So, we shall find the H.C.F. of 260 and 52 by Euclid’s division lemma.

∴ 260 =52 × 5 + 0.

Here, remainder = 0

⇒ The recent divisor is the H.C.F.

∴ H.C.F. of 260 and 52 is 52.

Thus, H.C.F. of 156, 208 and 260 is 52.

Hence, the minimum number ot buses = \(\frac{156}{52}+\frac{208}{52}+\frac{260}{52}\)

= 3 + 4 + 5

= 12

The minimum number ot buses = 12

Question 8. A sweetseller has 420 kaju burfis and 150 badam burfis. Mo wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. How many of these can be placed in each stack? How many stacks are formed?
Solution:

Given:

A sweetseller has 420 kaju burfis and 150 badam burfis. Mo wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray.

Maximum number of burfis in each stack = H.C.F. of 420 and 150

420 =2 × 2 × 3 × 5 × 7,

150 = 2 × 3 × 5 × 5

∴ H.C.F. =2 × 3 × 5

= 30

∴ Maximum number of burfis in each stack= 30

Also, number of stacks = \(\frac{420}{30}+\frac{150}{30}\)

= 14 = 5

= 19

Maximum number of burfis in each stack = 19

Fundamental Theorem Of Arithmetic

Every composite number can be uniquely expressed as a product of primes, except for the order in which these prime factors occurs.

For example., 143 =11 × 13

24 = 2 × 2 × 2 × 3

= 23 × 3

416 = 2 × 2 × 2 × 2 × 2 × 13

= 25 × 13

Real Numbers Difference Between Composite Number And Primary Number

Remember:

  1. 1 is not a prime number as it has no two different factors.
  2. 1 is not a composite number also because at least one factor other than 1 and the number must be there.
  3. 2 is the smallest prime number and it is the only even prime number also.
  4. The smallest even composite number is 4 while the smallest odd composite number is 9.

Fundamental Theorem Of Arithmetic Solved Examples

Question 1. Express each of the following as a product of prime factors:

  1. 1400
  2. 7650

Solution:

1. 1400

Real Numbers A Product Of Primary Factor 1

∴ 1400 = 2 × 2 × 2 × 5 × 5 × 7

Product of prime factors of 1400 = 2 × 2 × 2 × 5 × 5 × 7

2. 7650

Real Numbers A Product Of Primary Factor 2

∴ 7650 = 2 × 5 × 5 × 3 × 3 × 17

= 2 × 3 × 3 × 5 × 5 × 17

Product of prime factors of 7650 = 2 × 3 × 3 × 5 × 5 × 17

Question 2. Find the missing numbers in the following prime factorization.

Real Numbers The Missing Numbers In The Following Prime Factorisation

Solution:

The product of primes starts at the bottom of the factor tree and this product goes up to the top. The upper box, on 3 and 13 is filled by the product of 3 and 13, i.e., 39.

The upper next box on 2 and 39 will be filled by the product of 2 and 39, i.e., 78.

The topmost box on 2 and 78 will be filled by the product of 2 and 78, i.e., 156.

Real Numbers The Product Of Primes Factorisation

Question 3. Show that 5×11×17+17 is a composite number.
Solution:

5 × 11 × 17+ 17 = 17 × (5 × 11 + 1)

= 17 × (55 + 1)

= 17 × 56

∴ It has more than two factors including 1 and number itself.

Hence, it is a composite number.

Question 4. Find the H.C.F. and L.C.M. of 140 and 154 using the prime factorization method.
Solution:

Real Numbers HCF And LCM Of 140

Real Numbers HCF And LCM Of 154

Now,

Real Numbers A Product Of Prime Factorisation Method

∴ H.C.F = 2

L.C.M. = 2 x 2 x 5 x 7 x 11

= 1540

The H.C.F. and L.C.M. of 140 and 154 are 2 and 1540.

Question 5. Find the H.C.F. and L.C.M. of 12, 18, 24 by prime factorisation method.
Solution:

Real Numbers HCF And LCM Of 12

Real Numbers HCF And LCM Of 18

Real Numbers HCF And LCM Of 24

∴ 12 = 2 × 2 × 3

18 = 2 × 3 × 3

24 = 2 × 2 × 3 × 2

Now, H.C.F. = 2 × 3 = 6

and L.C.M. = 2 × 2 × 3 × 3 × 2

= 72

Question 6. Find the L.C.M. and H.C.F. of 36 and 48 and verify that H.C.F. x L.C.M. = product of the given two numbers.
Solution:

Real Numbers The LCM And HCF Of 36

Real Numbers The LCM And HCF Of 48

36 =2×2×3×3

48 =2×2×3×2×2

H.C.F. = 2×2×3=12

L.C.M. =2×2×3×3×2×2 = 144

Now H.C.F. x L.C.M. = 12 × 144

= 1728

and product of two numbers = 36 × 48

= 1 728

Hence, H.C.F. × L.C.M. = product of two numbers

An Important Result

Product of two given numbers = Product of their H.C.F. and L.C.M.

This result is true only for two numbers.

Question 7. The H.C.F of two numbers is 23 and their L.C.M. is 1449. If one number is 207, find the other number.
Solution:

Here, H.C.F = 23

L.C.M. = 1449

Now first no. x second no. = H.C.F. × L.C.M

⇒ Second no. = \(\frac{\text { H.C.F. } \times \text { L.C.M. }}{\text { first no. }}\)

⇒ \(\frac{23 \times 1449}{207}\)

= 161

The other number is 161

Question 8. Show that 12n cannot end with the digit 0 or 5 for any natural number n.
Solution:

Real Numbers Natural Number n

∴ 12 = 2×2×3 = 22×3

⇒ 12n = (22 x 3)n = 22n × 3n

∵ it has no term containing 5.

∴ no value of n eN for which 12″ ends with digit 0 or 5.

Hence Proved.

Question 9. On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm, and 45cm respectively. What is the minimum distance each should walk, so that each can cover the same distance in complete steps?
Solution:

Given:

On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm, and 45cm respectively.

We have to find a number (distance) which is divided by each number completely, which means → L.C.M.

We have to find the L.C.M. of 40 cm, 42 cm, and 45 cm to get the required distance.

Real Numbers The LCM Of 40 Cm

Real Numbers The LCM Of 42 Cm

Real Numbers The LCM Of 45

∴ 40 =2×2×2×5

42 =2 × 3 × 7

45 = 5 × 3 ×x 3

Now, L.C.M. = 2×2×2×5×3×7×3

= 2520

∴ The minimum distance each should walk = 2520 cm

Question 10. What is the smallest number which, when divided by 35, 56, and 91 leaves a remainder of 7 in each case?
Solution:

A number is divided by 3 numbers, which means → L.C.M. of 3 numbers

35 = 5 × 7

56=2×2×2×7

91 = 13 × 7

∴ L.C.M. = 5×7×2×2×2×13

= 3640

∴ The smallest number completely divisible by 35, 36, and 91 is 3640.

Hence, the smallest numbers which when divided by 35, 56, and 91 leave a remainder of 7 in each case will be 3640 + 7 = 3647

Question 11. Find the greatest number of 5 digits exactly divisible by 35, 56, and 91.
Solution:

A number is divided by 3 numbers.

It means → L.C.M. of 3 numbers.

35 = 5×7,56 = 2×2×2×7,

91 = 13 × 7

∴ L.C.M = \(\underbrace{5 \times 7}_{\text {1 as it is }} \times \underbrace{2 \times 2 \times 2}_{\begin{array}{c}
\text { from 2 } \\
\text { not taken yet }
\end{array}} \times \underbrace{13}_{\begin{array}{c}
\text { from 3 } \\
\text { not taken yet }
\end{array}}\)

= 3640

So, 3640 is the smallest number which is divided by all the 3 numbers 35, 56, and 91 completely. But we have to find the greatest number of 5 digits.

Greatest number of 5 digits = 99999

So, required number = 99999 – remainder when 99 is divided by 3640

= 99999- 1719 = 98280.

Alternatively: You can find the multiples of 3640 (L.C.M. of 35, 56 and 91) as

3640 × 1, 3640 × 2 3640 × 10 = 36400,…

3640 × 20 = 72800, 3640 × 25 = 91000,…

3640 × 27 = 98280 (It may be the largest 5 digit no.)

3640 × 28 = 101920 (No, it is 6 digit no.)

So, 98280 is the required number. But this is the time-talking method. So, avoid it)

Rational Numbers

The number which are in the form of \(\frac{p}{q}\) where p and q are integers and q ≠ 0 are called rational numbers.

Decimal Representation of a Rational Number

Consider the following examples:

1. \(\frac{1}{4}\)

= 0.25 4

2. \(\frac{2}{3}\)

In the above example 1, the decimal representation of the rational number \(\frac{1}{4}\) is terminating while in example 2, the decimal representation of the rational number \(\frac{2}{3}\) is non-terminating.

∴ Every rational number, when expressed in decimal form is expressible either in terminating or in non-terminating repeating decimal form.

Important Observation:

  1. A rational number \(\frac{p}{q}\) is a terminating decimal only if q can be written in the form of 2m × 5n for some non-negative integers in and n.
  2. A rational number — is a non-terminating repeating decimal if q cannot be written in the form of 2n × 5n.

For example:

  1. \(\frac{1}{14}=\frac{1}{2 \times 7}\) non-terminating after decimal, as denominator consists of 7 which cannot be the part of 2m x 5n.
  2. \(\frac{1}{3600}=\frac{1}{2^4 \times 3^2 \times 5^2}\) is non-terminating after decimal as denominator consists of 3’s which cannot be the part of \(2^m \times 5^n,\)
  3. \(\frac{91}{8750}=\frac{91}{2 \times 5^4 \times 7}\)
    1. Although it seems in first view that denominator is not the form \(2^m \times 5^n,\), after simplifying \(\frac{7 \times 13}{2 \times 5^4 \times 7}=\frac{13}{2 \times 5^4}\) we see that denominator is a part of \(2^m \times 5^n,\). So, \(\frac{91}{8750}\)is a terminating decimal.
    2. So, a rational number must be written in simplest form i.e., no factor other than 1 must be common to both the numerator and the denominator.
  4. \(\frac{3}{625}=\frac{3}{5^4}\)
    1. Although,it seems in the first view that the denominator is not in the form 2m x 5″, butit is our mistake.
    2. Actually, 5 =1×5 =2×5, which is of the form 2 x 5 . So,\(\frac{3}{625}\) is a terminating decimal

Irrational Numbers

The numbers, which when expressed in the decimal form are expressible as non-terminating and non-repeating decimals, are known as irrational numbers.

For example.,

  1. 2.101001000 is a non-terminating non-repeating decimal, so it is an irrational number.
  2. 1.767767776 is a non-terminating and non-repeating decimal, so it is an irrational number.

If x is a positive integer which is not a perfect square, then \(\sqrt{x}\) is irrational.

For example., \(\sqrt{2}, \sqrt{5}, \sqrt{7}\), etc., are irrational numbers.

Similarly \(\sqrt[3]{7}, \sqrt[4]{10}\), etc., are irrational numbers.

π is irrational and \(\frac{22}{7}\) is rational

Theorem: Let p is a prime number and V be a positive integer. If p divides a2, then show that p divides a.

Proof: We know that every positive integer can be expressed as a product of primes, not necessarily all distinct.

Let a = p1,p2,p3,……..pn

where, p1,p2,p3,……..pn are primes, not necessarily distinct.

∴ a2 = (p1,p2,p3,……..pn) (p1,p2,p3,……..pn)  = p21,p22,p23,……..p2n

Now, p divides a2

⇒ p is a prime factor of a2.

⇒ P is one of P1,p2,P3 …….. P„-

⇒ p divides a.

Irrational Numbers Solved Examples

Question 1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

  1. \(\frac{12}{125}\)
  2. \(\frac{7}{1600}\)

Solution:

1. \(\frac{12}{125}=\frac{12}{5 \times 5 \times 5}=\frac{12}{5^3}=\frac{12}{2^0 \times 5^3}\)

Now, the denominator is in the form of 2m x 5”.

∴ \(\frac{12}{125}\) is a terminating decimal.

2. \(\frac{7}{1600}\)

⇒ \(\frac{7}{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5}\)

⇒ \(\frac{7}{2^6 \times 5^2}\)

Now, the denominator is in the form of 2m x 5″.

∴ \(\frac{7}{1600}\) is a terminating decimal.

Question 2. Show that :

  1. \(\frac{3}{250}\)
  2. \(\frac{11}{50}\)

are terminating decimals. Express each of them in decimal form without actual division (long division).

Solution:

1. \(\frac{3}{250}=\frac{3}{2 \times 5 \times 5 \times 5}\)

\(=\frac{3}{2^1 \times 5^3}\)

Now, the denominator is in the form of 2m × 5n

∴ \(\frac{3}{250}\) is a terminating decimal.

Agan, \(\frac{3}{250}\)

⇒ \(\frac{3}{2^1 \times 5^3}\)

⇒ \(\frac{3 \times 2^2}{2^3 \times 5^3}\)

⇒ \(\frac{12}{10^3}]\)

= 0.012

2. \(\frac{11}{50}\)

⇒ \(\frac{11}{2 \times 5 \times 5}\)

⇒ \(\frac{11}{2^1 \times 5^2}\)

Now, the denominator is in the form of 2m × 5n

∴ \(\frac{11}{50}\)

Again, \(\frac{11}{50}\)

\(\frac{11}{2^1 \times 5^2}\) \(\frac{11 \times 2}{2^2 \times 5^2}\) \(\frac{22}{10^2}\)

= 0.22

Question 3.  Show that each of the following are non-terminating repeating decimal:

  1. \(\frac{5}{12}\)
  2. \(\frac{7}{75}\)

Solution:

1. \(\frac{5}{12}\)

⇒ \(\frac{5}{2 \times 2 \times 3}\)

⇒ \(\frac{5}{2^2 \times 3}\)

the denominator 22 x 3 is not in the form of 2m x 5n

∴ it is non-terminating repeating decimal.

2. \(\frac{7}{25}\)

⇒ \(\frac{7}{3 \times 5 \times 5}\)

⇒ \(\frac{7}{3 \times 5^2}\)

the denominator 3 x 52 is not in the form of 2n x 5m

∴ it is non-terminating repeating decimal.

Question 4. The decimal expansion of the rational number \(\frac{43}{2^4 \times 5^3}\) will terminate after how many places of decimals?
Solution:

\(\frac{43}{2^4 \times 5^3}\) \(\frac{43 \times 5}{2^4 \times 5^4}\) \(\frac{215}{10^4}\)

= 0.0215

∴ it will terminate after 4 places of decimals.

Question 5. Express each of the following in the simplest form:

  1. \(0. \overline{6}\)
  2. \(3. \overline{3}\)

Solution:

1. Let x = \(0 . \overline{6}\)

⇒ x = 0.666….. (1)

⇒ 10x = 6.666….. (2)

Subtracting equation (1) from (2), we get

9x = 6

⇒ \(x=\frac{6}{9}=\frac{2}{3}\)

⇒ \(0. \overline{6}=\frac{2}{3}\)

2. Let x = \(3 . \overline{3}\)

⇒ x = 3.333…(1)

⇒ 10x = 33.333…(2)

Subtracting equation (1) from (2), we get

9x = 30

⇒ \(x=\frac{30}{9}=\frac{10}{3}\)

⇒ \(3. \overline{3}=\frac{10}{3}\)

Question 6. Express each of the following in the simplest form :

Let x = \(0. \overline{36}\)

⇒ \(1. \overline{046}\)

⇒ \(Let x=0. \overline{36}\)

⇒\(x=0.363636 \ldots . . .\)

⇒ \(100 x=36.363636 \ldots . . \)

Subtracting equation (1) from (2), we get

⇒ \(99 x =36\)

⇒ \(x =\frac{36}{99}=\frac{4}{11}\)

⇒ \( 0 . \overline{36}=\frac{4}{11}\)

2. Let \(x=1. \overline{046}$

⇒ [latex] x  =1.046046046 \ldots\)

⇒ \(1000 x  =1046.046046046 \ldots \)

Subtracting equation (1) from (2), we get

⇒ \(999 x =1045\)

⇒ \(x=\frac{1045}{999}\)

⇒ \(1. \overline{046}\)

⇒ \(\frac{1045}{999}\)

Question 7. A rational number in its recurring decimal expansion is 327.7081. What can you say about the prime factors of q, when this number is expressed in the form \(\frac{p}{q}\)? Give <l reasons.
Solution:

Given:

A rational number in its recurring decimal expansion is 327.7081.

Let x = \(327.7 \overline{081}\)

⇒ x =327.7081081081…..

10x =3277.081081081…..

10000 A- =3277081.081081081…..

On subtraction, 9990x =3273804

∴ x = \(\frac{3273804}{9990}=\frac{60626}{185}\)

which is of the form =\(\frac{p}{q}\)

Now, q= 185 = 5 × 37,

which cannot be written in the form 2m x 5n.

∴ It is a non-terminating and repeating decimal.

Question 8. Prove that \(\sqrt{2}\) is irrational.
Solution:

Let, if possible. \(\sqrt{2}\) be rational and its simplest form be \(\frac{a}{b}\)

Then a and b are integers and have no common factor other than 1 and b ≠ 0

Now, \(\sqrt{2}=\frac{a}{b}\)

⇒ \(2=\frac{a^2}{b^2}\)

⇒ \(a^2=2 b^2\)

As 2b2 is divisible by 2.

∴ a2 is divisible by 2.

⇒ a is divisible by 2.

Let a = 2c, for some integer c.

∴ From equation (1)

(2c)2 = 2b2

⇒ b2 = 2c2

But 2c2 is divisible by 2.

∴ b2 is divisible by 2.

⇒ b is divisible by 2.

Let b = 2d, for some integer d.

Thus, 2 is a common factor of a and b both.

But it contradicts the fact that a and b have no common factor other than 1.

So, our supposition is wrong.

Hence, \(\sqrt{2}\) is irrational.

Hence Proved.

Question 9. If p is a prime number, then prove that \(\sqrt{p}\) is irrational. (Treat this result as a theorem)
Solution:

Let p be a prime number and if possible let \(\sqrt{p}\) be irrational.

Let the simplest form of \(\sqrt{p} \text { be } \frac{a}{b} \text {. }\)

Then a and b are integers and having no common factors other than 1 and b ≠ 0.

Now, \( \sqrt{p}=\frac{u}{b}\)

⇒ \(p=\frac{a^2}{b^2}\)

⇒ \(a^2=p b^2\)

As pb2 is divisible by p.

∴ a2 is divisible by p.

⇒ a is divisible by p.

Let a = pc for some integer c.

From equation (1)

⇒ \((p c)^2=p b^2\)

⇒ \(b^2=p c^2\)

But pc2 is divisible by p.

∴ b2 is divisible by p.

⇒ b is divisible by p.

Let b = pd for some integer d.

Thus, p is a common factor of both a and b.

But it contradicts the fact that a and b have no common factor other than 1.

So, our supposition is wrong.

Hence, \(\sqrt{2}\) is irrational.

Question 10. Show that \((2+\sqrt{3})\) is an irrational number.
Solution:

Let, if possible \((2+\sqrt{3})\) is rational

then, \(2+\sqrt{3}=\frac{a}{b} \text { (say) }\)

where a and b are integers and b ≠ 0

⇒ \(\sqrt{3}=\frac{a}{b}-2\)

⇒ \(\sqrt{3}=\frac{a-2 b}{b}\)

a and b are integers

∴ a – 2b is also an integer.

⇒ \(\frac{a-2 b}{b}\) is rational

Now, L.H.S of equation (1) is the square root of a prime number. So, it is irrational and R.H.S is rational.

Which is a contradiction because a rational number and an irrational number can never be equal.

So, our supposition i.e., \((2+\sqrt{3})\) is rational, is wrong.

Question 11. Prove that \(5 \sqrt{7}\) is irrational
Solution:

Let if possible \(5 \sqrt{7}\) is rational.

Now, \(5 \sqrt{7}\)is rational and \(\frac{1}{5}\) is rational.

We know that the product of two rational numbers is rational.

∴ \(5 \sqrt{7} \times \frac{1}{5} \text { is rational. }\) is rational.

⇒ \(\sqrt{7}\)is rational.

But square root of a prime number is always an irrational number. This contradicts the fact because an irrational number cannot be equal to a rational number.

So, our supposition is wrong.

Hence, \(5 \sqrt{7}\) is irrational.

Question 12. A rational number in its decimal is 327.7081. What can you say about the prime factor of q. when this number is expressed in the form \(\frac{p}{q}\)? Give reasons.
Solution:

Given:

A rational number in its decimal is 327.7081.

The decimal expansion is 327.7081.

∴ It is a rational number.

Now, 327.7081 = \(\frac{3277081}{10000}\)

⇒ \(\frac{3277081}{2^4 \times 5^4}\)

⇒ \(\frac{p}{q}\)

Here q = \(2^4 \times 5^4\) which is in the form of \(2^m \times 5^n\)

∴ The prime factors of q are 2 and 5

Real Numbers Exercise 1.1

Question 1. Use Euclid’s division algorithm to find the H.C.F. of :

  1. 135 and 225
  2. 196 and 38220
  3. 867 and 255

Solution:

1. Given numbers 135 and 225

Real Numbers The HCF Of 135 And 225

Here, 225 > 135

∴ 225 = 135 × 1 + 90

Remainder = 90 ≠ 0

135 = 90 × 1 + 45

Remainder = 45 ≠ 0

90 = 45 × 2 + 0

Remainder = 0

∵ The remainder is zero and the divisor is 45

∴ H.C.F. = 45

The H.C.F. of 135 and 225 = 45

2. Given numbers 196 and 38220

Real Numbers The HCF Of 196 And 38220

38220 > 196

∴ 38220 = 196 × 195 + 0

Remainder = 0

∵ The remainder is zero and the divisor is 196.

∴ H.C.F. = 196

The H.C.F. of 196 and 38220 is 196

3. Given numbers 867 and 255

Real Numbers HCF Of 255 And 869

867 > 255

∴ 867 = 255 x 3 + 102

Remainder = 102 ≠ 0

255 = 102 × 2 + 51

Remainder = 51 ≠ 0

102 = 51 × 2 + 0

Remainder = 0

∵ The remainder is zero and the divisor is 51.

∴ H.C.F. =51

The H.C.F. of 867 and 255 is 51

Question 2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solution:

Let ‘a’ be an odd positive integer.

From Euclid’s division algorithm,

Let q be the quotient and the remainder when a is divided by 6.

∴ a = 6q +r

where, r = 0,1,2,3,4,5

Now, a = 6q+ 0 or a = 6q + 1 or a = 6q + 2 or a = 6q + 3 or a = 6q + 4 or a = 6q + 5 but 6q + 0, 6q + 2, 6q + 4 are even numbers.

∴ a = 6q + 1 or 6q + 3 or 6q + 5

Therefore, any positive odd integer is of the form 6q+1 or 6q + 3 or 6q + 5 where q is some integer.

Question 3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solution :

Given:

An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns.

The maximum number of columns will be the H.C.F. of the number of army contingent of 616 members and the number of army band of 32 members.

Here, 616> 32

Real Numbers The HCF Of 32 And 616

Now, 616 = 32 × 19 + 8

Remainder = 8≠0

32 = 8 × 4 + 0

Remainder = 0

∵ The remainder is zero and the divisor is 8.

∴ H.C.F. = 8

Therefore, an army can march in 8 columns.

Question 4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Solution :

Let ‘a’ be a positive integer.

From Euclid’s division algorithm,

Let q be the quotient and r the remainder when a is divided by 3,

∴ a = 3q + r where r = 0,1,2

⇒ a = 3q + 0 or a = 3q + 1 or a = 3q + 2

Now a = 3q

⇒ a2 = (3q)2

= 9q2 = 3(3q2)

= 3m

where, m = 3q2 is an integer.

Again, a = 3q + 1

⇒ a2 = (3 q+1)2 = 9q2+ 6q + 1

= 3(3q2+2q) +1 = 3m + 1

where, m = 3q2 + 2q is an integer and

a = 3q +2

⇒  a2 = (3 q + 2)2

= 9q2 + 12q + 4

= 3(3q2 + 4q + 1) + 1

= 3m + 1

where, m = 3q2 + 4q + 1 is an integer.

So, a2 = 3m or 3m + 1.

The square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Hence Proved.

Question 5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.
Solution:

Let ‘a’ be a positive integer,

From Euclid’s division algorithm,

Let q be the quotient and r, be the remainder when a is divided by 3.

∴ a = 3q + r

where r = 0, 1, 2

⇒ a = 3q + 0 or a = 3q + 1 or a = 3q + 2

Now, a = 3

⇒ a3 = (3q)3 = 27q3 = 9(3q3) = 9m.

where = 3q3 is an integer.

Again, a = 3q + 1

⇒ \( a=3 q+1\quad a^3=(3 q+1)^3\)

⇒ \(27 q^3+3(3 q)(1)(3 q+1)+1\)

⇒ \(27 q^3+9 q(3 q+1)+1\)

⇒ \(9\left[3 q^3+3 q^2+q\right]+1=9 m+1\)

where, m = 3q3 + 3q2+q is an integer.

Again, a = 3q + 2

⇒ \( a^3=(3 q+2)^3\)

⇒ \(27 q^3+3(3 q)(2)(3 q+2)+8\)

⇒ \(9\left[3 q^3+6 q^2+4 q\right]+8\)

= 9m+8

Where \(m=3 q^3+6 q^2+4 q \text { is an integer. }\)

∴ a3 = 9m or 9m + 1 or 9m + 8

The cube of any positive integer is of the form 9m or 9m + 1 or 9m + 8.

Real Numbers Exercise 1.2

Question 1. Express each number as a product of its prime factors :

  1. 140
  2. 156
  3. 3825
  4. 5005
  5. 7429

Solution:

1. 140 = 2 × 2 × 5 × 7

= 22 × 5 × 7

Real Numbers Product Of Prime Factor 1

2.  156 = 2 × 2 ×3 × 13

= 22 × 3 × 13

Real Numbers Product Of Prime Factor 2

3. 3825 = 3 × 3 × 5 × 5 × 17

= 32 × 52 × 17

Real Numbers Product Of Prime Factor 3

4. 5005 = 5 × 7 × 11 × 13

Real Numbers Product Of Prime Factor 4

5. 7429 = 17 × 19 × 23

Real Numbers Product Of Prime Factor 5.

Question 2. Find the L.C.M. and H.C.F. of the following pairs of integers and verify that L.C.M. x H.C.F. = product of the two numbers.

  1. 26 and 91
  2. 510 and 92
  3. 336 and 54

Solution:

1. 26 = 2 × 13

91 = 13 × 7

H.C.F. = 13

L.C.M. = 2 × 13 × 7

= 182

Product of numbers = 26 x 91 = 2366

H.C.F. x L.C.M. = 13 x 182 = 2366

∴ Product of numbers = H.C.F. × L.C.M.

2. 510 = 2 × 3 × 5 × 17

92 = 2 × 2 × 23

H.C.F. = 2

L.C.M. =2 × 3 × 5 × 17 × 2 × 23

= 23460

Now, product of numbers = 510 × 92

= 46920

H.C.F. x L.C.M. = 2 x 23460 = 46920

∴ Product of numbers = H.C.F. x L.C.M.

336 = 2x2x2x2x3x7

54 = 2x3x3x3

H.C.F. =2×3 = 6

L.C.M. = 2x2x2x2x3x7x3x3

= 3024

Now, product of numbers = 336 x 54 = 18144

H.C.F. x L.C.M. = 6 x 3024 = 18144

∴ Product of numbers

H.C.F. x L.C.M

Question 3. Find the L.C.M and H.C.F of the following integers by applying the prime factorization method:

  1. 12. 15 and 21
  2. 17, 23 and 29
  3. 8. 9 and 25

Solution:

1. 12 = 2 × 2 × 3

15 =3 × 5

21 =3 × 7

∴ H.C.F. = 3

L.C.M. = 2×2×3×5×7

= 420

The L.C.M and H.C.F of 12. 15 and 21 is 420 and 3.

2. 17 = 17 × 1

23 = 1 × 23

29 = 1 × 29

∴ H.C.F. = 1

L.C.M. = 1 7 × 23 × 29 = 11339

The L.C.M and H.C.F of 17, 23 and 29 is 11339 and 1.

3. 8 = 2 × 2 × 2

9 = 3 × 3

25 =5 × 5

∴ H.C.F. = 1

L.C.M. = 2×2×2×3×3×5×5

= 1800

The L.C.M and H.C.F of 8. 9 and 25 is 1800 and 1.

Question 4. Given that H.C.F. (306, 657) = 9, find L.C.M. (306, 657).
Solution:

H.C.F. (306, 657) = 9

H.C.F. × L.C.M. = Product of numbers

9 × L.C.M. = 306 x 657

306 × 657

⇒ L.C.M. = \(\frac{306 \times 657}{9}\)

= 34 × 657

= 22338

L.C.M. of (306, 657) = 22338

Question 5. Check whether 6″ can end with the digit 0 for any natural number n.
Solution:

6 = 2 × 3

6n = (2 x 3)n

5 is not a factor of 6.

Real Numbers Exercise 1.3

Question 1. Prove that \(\sqrt{5}\) is irrational.
Solution:

Let \(\sqrt{5}\) is a rational number.

Let \(\sqrt{5}\) = where\(\frac{a}{b}\) and a andb are integers,

which have no common prime factors other than 1.

Now, \(\) = \(\frac{a}{b}\) ….(1)

⇒ a2 is divisible by 5.

⇒ a is divisible by 5

Let a = 5c

⇒ a2 = 25c2

⇒ 5b2 = 25c2 (From equation (1) )

⇒ b2 = 5c2

Now, 5c2 is divisible by 5.

⇒ b2 is divisible by 5.

⇒ b is divisible by 5

∴ 5 is a common factor of a and b.

It is opposite to our assumption.

∴ Our assumption is wrong.

i.e., \(\sqrt{5}\) is an irrational number.

Hence proved.

Question 2. Prove that \(3+2 \sqrt{5}\) is irrational:
Solution:

Let \(3+2 \sqrt{5}\) is a rational number

Let \(\) = \(\) where q* 0 and p and q are positive integers.

⇒ \(2 \sqrt{5}=\frac{p}{q}-3\)

⇒ \(\frac{p-3 q}{q}\)

⇒ \(\sqrt{5}=\frac{p-3 q}{2 q}\)

∵ p and q are integers and q≠0.

∴ \(\frac{p-3 q}{2 q}\) is a rational number.

From question 1,\(\sqrt{5}\) is an irrational.

Now, an irrational number and a rational number are equal.

Which is not possible.

So, it is opposite to our assumption.

∴ Our assumption is wrong.

i.e., \(3+2 \sqrt{5}\) is an irrational number.

Question 3. Prove that the following are irrationals :

  1. \(\frac{1}{\sqrt{2}}\)
  2. \(\frac{7}{\sqrt{5}}\)
  3. \(6+\sqrt{2}\)

Solution:

1. \(\frac{1}{\sqrt{2}}\)

Let \(\frac{1}{\sqrt{2}}\) is a rational number

Let \(\frac{1}{\sqrt{2}}=\frac{p}{q}\)= where q≠0 andp and q are integers.

⇒ \(q=\sqrt{2} p \Rightarrow q^2=2 p^2\) ……(1)

Now, 2p2 is divisible by 2.

⇒ q2 is divisible by 2.

⇒ q is divisible by 2.

Let q = 2r

⇒ q2 = 4r2

⇒ 2p2 = 4r2    (from equation 1)

⇒ p2 = 2r2

Now, 2r2 is divisible by 2.

⇒ p2 is divisible by 2.

⇒ p is divisible by 2.

∴ 2 is a common factor of p and q, which is opposite to our assumption.

So, our assumption is wrong i.e., \(\frac{1}{\sqrt{2}}\) is an So, our assumption is wrong i.e., irrational number.

2.  Let \(7 \sqrt{5}\) is a rational number.

Let \(7 \sqrt{5}=\frac{p}{q}\) where q*0 andp andq are integers.

⇒ \(\sqrt{5}=\frac{p}{7 q}\)

∵ p and are integers and 4 * 0

∴ \(\frac{p}{7 q}\) is a rational number

From question 1,\(\sqrt{5}\) is an irrational number. Now, an irrational number and a rational number are equal, which is impossible. So, it is opposite to our assumption:

∴ Our assumption is wrong, i.e., \(7 \sqrt{5}\) is an irrational number.

3.  Let \(6+\sqrt{2}\) is a rational number.

Let \(6+\sqrt{2}=\frac{p}{q}\) where q * 0 and p and q are integers.

⇒ \(\sqrt{2}=\frac{p}{q}-6\)

Now,  \(\frac{p}{q}-6\)  is rational number and \(\sqrt{2}\) is an irrational number.

Now, an irrational number and a rational number are equal.

Which is not possible.

So, it is opposite to our assumption.

∴ Our assumption is wrong.

Therefore, \(6+\sqrt{2}\) is an irrational number.

Real Numbers Exercise 1.4

Question 1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion :

  1. \(\frac{13}{3125}\)
  2. \(\frac{17}{8}\)
  3. \(\frac{64}{455}\)
  4. \(\frac{15}{1600}\)
  5. \(\frac{29}{343}\)
  6. \(\frac{23}{2^3 5^2}\)
  7. \(\frac{129}{2^2 5^7 7^5}\)
  8. \(\frac{6}{15}\)
  9. \(\frac{35}{50}\)
  10. \(\frac{77}{210}\)

Solution:

1. \(\frac{13}{3125}\)

⇒ \(\frac{13}{3125}\)

⇒ \(=\frac{13}{5 \times 5 \times 5 \times 5 \times 5}\)

⇒ \(=\frac{13}{5^5}\)

⇒ \(\frac{13}{2^0 \times 5^5}\)

Its denominator is 2° × 55 whose prime factors are 2 and 5 only.

Therefore, the decimal expansion of \(\frac{13}{3125}\) is terminating.

2. \(\frac{17}{8}\)

⇒ \(\frac{17}{2 \times 2 \times 2}\)

⇒ \(=\frac{17}{2^3 \times 5^0}\)

Its denominator is 23 × 5° whose prime factors are 2 and 5 only.

Therefore, the decimal expansion of \(\) is terminating

3. \(\frac{64}{455}\)

⇒ \(\frac{64}{5 \times 7 \times 13}\)

Its denominator has the prime factors 7 and 13 other than 5, which is not in form 2m × 5n. So the decimal expansion of \(\frac{64}{455}\) is non-terminating and repeating.

4. \(\frac{15}{1600}\)

⇒ \(=\frac{3 \times 5}{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5}\)

⇒ \(\frac{3}{2^6 \times 5^1}\)

Its denominator is 26 × 51 whose prime factors are 2 and 5 only.

Therefore, the decimal expansion of \(\frac{15}{1600}\) is terminating.

5. \(\frac{29}{343}\)

⇒ \(\frac{29}{7 \times 7 \times 7}\)

⇒ \(\frac{29}{7^3}\)

7 is a prime factor of its denominator which is not in the form 2m x 5n.

Therefore, the decimal expansion of is \(\frac{29}{343}\) non-terminating and repeating.

6. \(\frac{23}{2^3 \cdot 5^2}\)

Its denominator is 23. 52 whose prime factors are 2 and 5 only.

Therefore, the decimal expansion of \(\frac{23}{2^3 \cdot 5^2}\) is terminating.

7. \(\frac{129}{2^2 \cdot 5^7 \cdot 7^5}\)

Its denominator is 2n.57. 75 which is not in the form 2m × 5n.

Therefore, the decimal expansion of \(\frac{129}{2^2 \cdot 5^7 \cdot 7^5}\) is non-terminating and repeating.

8. \(\frac{6}{15}\)

⇒ \(\frac{3 \times 2}{3 \times 5}\)

⇒ \(\frac{2}{5}\)

Its denominator is 5 whose prime factor is 5 only.

Therefore, the decimal expansion of \(=\frac{6}{15}\) is terminating.

9. \(\frac{35}{50}\)

⇒ \(\frac{5 \times 7}{2 \times 5 \times 5}\)

⇒ \(\frac{7}{2 \times 5}\)

Its denominator is 2 x 5 whose prime factors are 2 and 5 only.

Therefore, the decimal expansion of \(\frac{35}{50}\) is terminating.

10. \(\frac{77}{210}\)

⇒ \(\frac{7 \times 11}{2 \times 5 \times 3 \times 7}\)

⇒ \(\frac{11}{2 \times 5 \times 3}\)

Its denominator is 2×5×3 which is not in the form 2nx5n

Therefore, the decimal expansion of \(\frac{77}{210}\) is non-terminating and repeating.

Question 2. Write down the decimal expansions of those rational numbers above which have terminating decimal expansions.
Solution:

1. \(\frac{13}{3125}\)

⇒ \(\frac{13}{5^5}\)

⇒ \(\frac{13 \times 2^5}{5^5 \times 2^5}\)

⇒ \(\frac{13 \times 32}{(5 \times 2)^5}\)

⇒ \(\frac{416}{10^5}\)

=0.00416

\(\frac{13}{3125}\) =0.00416

2. \(\frac{17}{8}\)

⇒ \(\frac{17}{2^3}\)

⇒ \(\frac{17 \times 5^3}{2^3 \times 5^3}\)

⇒ \(\frac{17 \times 125}{(2 \times 5)^3}\)

\(\frac{17}{8}\) ⇒ \(\frac{17 \times 125}{(2 \times 5)^3}\)

3. \(\frac{15}{1600}\)

⇒ \(\frac{3 \times 5}{2^6 \times 5^2}\)

⇒ \(\frac{3 \times 5 \times 5^4}{2^6 \times 5^6}\)

⇒ \(\frac{15 \times 625}{(2 \times 5)^6}\)

⇒ \(\frac{9375}{10^6}\)

=0.009375

\(\frac{15}{1600}\) =0.009375

4. \(\frac{23}{2^3 5^2}\)

⇒ \(\frac{23 \times 5}{2^3 \times 5^3}\)

⇒ \(\frac{115}{(2 \times 5)^3}\)

⇒ \(\frac{115}{10^3}\)

=0.115

\(\frac{23}{2^3 5^2}\) =0.115

5. \(\frac{6}{15}\)

⇒ \(\frac{2 \times 3}{3 \times 5}\)

⇒ \(\frac{2}{5}\)

⇒ \(\frac{2 \times 2}{5 \times 2}\)

⇒ \(\frac{4}{10}=0.4\)

\(\frac{6}{15}\) ⇒ \(\frac{4}{10}=0.4\)

6. \(\frac{35}{50}\)

⇒ \(\frac{5 \times 7}{5 \times 5 \times 2}\)

⇒ \(\frac{5 \times 7 \times 2}{5^2 \times 2^2}\)

⇒ \(\frac{70}{(10)^2}\)

=0.7

\(\frac{35}{50}\) =0.7

Question 3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they rational’ and of the form \(\frac{p}{q}\) what can you say about the prime factors of q?

  1. 43.123456789
  2. 0.120120012000120000…
  3. \(43. \overline{123456789}\)

Solution:

1. 43.123456789 = \(\frac{43123456789}{1000000000}\)

which is in the form \(\frac{p}{q}\) So, it is a prime number

1000000000 has prime factors 2 and 5.

2. 0.120120012000120000…

Its decimal expansion is non-terminating and non-repeating. It cannot be expressed in the form \(\frac{p}{q}\)

So, it is an irrational number.

3. \(43. \overline{123456789}\)

Its decimal expansion is non-terminating and repeating.

So, it is a rational number.

The given number can be expressed in the form \(\frac{p}{q}\)

It has some other prime factor except 2 or 5.

Multiple Choice Questions And Answers

Question 1. The decimal expansion of the rational number \(\frac{17}{2^2 \cdot 5}\)ends after the following decimal places :

  1. 1
  2. 2
  3. 3
  4. 4

Answer: 2. 2

Question 2. For some integer m, every odd integer is of the following form :

  1. m
  2. 2m
  3. m+1
  4. 2m+1

Answer: 4. 2m+1

Question 3. For some integer m, every even integer is of the following form :

  1. m
  2. 2m
  3. m+1
  4. 2m+1

Answer: 2. 2m

Question 4. 4-. The largest number from which 57 and 67 divided, leaving the remainders 5 and 7 respectively, is:

  1. 5
  2. 8
  3. 10
  4. 11

Answer: 3. 10

Question 5. The sum of a non-zero rational number and an irrational number is:

  1. Rational
  2. Irrational
  3. Zero
  4. None of these

Answer: 2. Irrational

Question 6. If a = x2y and b =xy2 then HCF (a, b) is

  1. x
  2. y
  3. xy
  4. x2y2

Answer: 3. xy

Question 7. The decimal expansion of \(\frac{3721}{625}\) ends after the following decimal places:

  1. 1
  2. 2
  3. 3
  4. 4

Answer: 4. 4

Question 8.  Given that LCM (132, 288) = 3168 then HCF (132, 288) is :

  1. 288
  2. 132
  3. 48
  4. 12

Answer: 4. 12

Question 9. The sum of exponents of prime factors in the prime factorization of the number 144 is :

  1. 3
  2. 4
  3. 5
  4. 6

Answer: 4. 6

Question 10. The prime number is:

  1. 0
  2. 1
  3. 2
  4. 8

Answer: 3. 2

Question 11. If the L.C.M. of 26, 156 is 156, then the value of HCF is:

  1. 156
  2. 26
  3. 13
  4. 6

Answer: 2. 26