Well-designed AP 10th Class Maths Textbook Solutions Chapter 5 Arithmetic Progressions Exercise 5.2 offers step-by-step explanations to help students understand problem-solving strategies.

## Arithmetic Progressions Class 10 Exercise 5.2 Solutions – 10th Class Maths 5.2 Exercise Solutions

Question 1.

Fill in the blanks in the following table, given that a is the first term, d the common difference and a_{n} is the n^{th} term of the AP:

Solution:

Question 2.

Choose the correct choice in the following and justify:

i) 30th term of the AP: 10, 7, 4,…,is

A) 97

B) 77

C) -77

D) -87

Solution:

C) -77

Given A.P is 10, 7, 4, … and n = 30

a = 10, d = a_{2} – a_{1} = 7 – 10 = -3

a_{n} = a + (n – 1)d

a_{30} = 10 + (30 – 1)(-3) = 10 + 29 × (-3)

a_{30} = 10 – 87 = -77

ii) 11^{th} term of A.P: -3, –\(\frac{1}{2}\),2,……is

A) 28

B) 22

C) -38

D) -48

Solution:

B) 22

Given A.P is : -3, –\(\frac{1}{2}\), 2,…….. and n = 11

a = -3, d = a_{2} – a_{1}

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

a_{n} = a + (n – 1)d

a_{11} = -3 + (11 – 1)\(\frac{5}{2}\)

a_{11} = -3 + 25 = 22

Question 3.

In the following APs, find the missing terms in the boxes:

Solution:

i) Given 2, ____, 26

a_{1} = 2, a_{3} = 26

d = a_{2} – a_{1} = a_{3} – a_{2}

____ – 2 = 26 – ____

___ + _____ = 26 + 2

2 ∙ ___ = 28

______ = \(\frac{28}{2}\) = 14

Therefore A.P. is 2, __14__, 26.

ii)

⇒ a + d = 13

a + (-5) = 13

a – 5 = 13

a = 13 + 5 = 18

a, 13, a + 2d, 3

18, 13, 18 + 2(-5), 3

∴ A.P is __18__, 13, __8__, 3

iii) Given 5, a + d, a + 2d, 9\(\frac{1}{2}\)

iv) Given -4, a_{2}, a_{3}, a_{4}, a_{5}, 6

a_{1} = a = -4, a + 5d = 6

-4 + 5d = 6

5d = 6 + 4 = 10

d = \(\frac{10}{5}\) = 2

a_{2} = a_{1} + d = -4 + 2 = -2

a_{3} = a_{1} + 2d = -4 + 2(+2) = -4 + 4 = 0

a_{4} = a_{1} + 3d = -4 + 3(2) = -4 + 6 = 2

a_{5} = a_{1} + 4d = -4 + 4(2) = -4 + 8 = 4

∴ A.P is -4, __-2__, __0__, __2__, __4__, 6.

v) Given a_{1}, 38, a__3__, a__4__, a__5__, -22

a, a + d, a + 2d, a + 3d, a + 4d, a + 5d

a + d = 38 → (1)

a + 5d = -22 → (2)

(2) – (1)

⇒

Put d = -15 in (1)

⇒ a + d = 38

a + (-15) = 38

a = 38 + 15 = 53

∴ a = 53

a, a + d, a+ 2d, a+ 3d, a + 4d, a + 5d

= 53, 38, 53 + 2(-15), 53 + 3(-15), 53 + 4(-15), -22

53, 38, 53-30, 53-45, 53-60, -22

∴ A.P is __53__, 38, __23__, __8__, __-7__, -22.

Question 4.

Which term of the AP : 3, 8, 13, 18,……, is 78?

Solution:

Given A.P is 3, 8, 13, 18, … and a = 78

a = 3, d = a_{2} – a_{1} = 8 – 3 = 5

We have, a = a + (n – 1)d

⇒ 3 + (n – 1)5 = 78

⇒ (n – 1)5 = 78 – 3 = 75

⇒ n – 1 = \(\frac{75}{5}\) = 15

∴ n = 15 + 1 = 16

∴ 16^{th} term is 78.

Question 5.

Find the number of terms in each of the following APs:

i) 7, 13, 19, ………, 205

ii) 18, 15\(\frac{1}{2}\), 13,……,-47

Solution:

i) Given A.P is 7, 13, 19, …, 205

a = 7, d = 13 – 7 = 6, last term l = 205

l = a + (n – 1)d = 205

7 + (n – 1)6 = 205

⇒ (n – 1)6 = 205 – 7 = 198

n – 1 = \(\frac{198}{6}\) = 33

Therefore, n = 33 + 1 =34

So, there are 34 terms in the given A.P.

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

a = 18, last term l = -47

d = a_{2} – a_{1} = 15\(\frac{1}{2}\) – 18 = \(\frac{31}{2}\) – 18

⇒ d = \(\frac{-5}{2}\)

Last term l = a + (n – 1) d

Therefore, n = 26 + 1 = 27

So, there are 27 terms in the given A.P.

Question 6.

Check whether – 150 is a term of the AP : 11, 8, 5, 2,…….

Solution:

Given A.P is 11, 8, 5, 2, …,-150

a = 11, d = a_{2} – a_{1} = 8 – 11 = -3,

Let, last term l = -150

Last term l = a + (n – 1) d

⇒ 11 + (n – 1) (-3) = -150

⇒ (n – 1)(-3) = -150 – 11 = -161

⇒ n = \(\frac{161}{3}+\frac{1}{1}\) = \(\frac{164}{3}\)

Therefore, n = \(\frac{164}{3}\)

n = \(\frac{164}{3}\) is not be a positive integer.

So, -150 is not a term of the given A.P.

Question 7.

Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Solution:

Given A.P a_{11} = 38 and a_{16} = 73

a_{11} = a + 10d = 38

a_{16} = a + 15d = 73

(2) – (1) ⇒

Put d = -7 in (1) ⇒ a_{11} = a + 10d = 38

⇒ a + 10(7) = 38

⇒ a = 38 – 70 = -32

31^{st} term is a_{31} = a + 30d = -32 + 30(7)

a_{31} = -32 + 210 = 178

Therefore, 31^{st} term in A.P is 178.

Question 8.

An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution:

Given 3^{rd} term of A.P: a_{3} = 12,

last term l = 106

Number of terms n = 50

a_{3} = a + 2d = 12 → (1)

l = a + (n – 1)dz

⇒ a + (50 – 1)d = 106

⇒ a + 49d = 106 → (2)

(2) – (1)

Put d = 2 in (1) ⇒ a + 2d = 12

⇒ a + 2(2) = 12

⇒ a = 12 – 4 = 8

29^{th} term is a_{29} = a + 28d

= 8 + 28(2)

= 856 = 64

Therefore, 29^{th} term is 64.

Question 9.

If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?

Solution:

Given 3^{rd} term of an A.P is a_{3} = 4

and 9^{th} term is a_{9} = -8

a_{3} = a + 2d = 4 → (1)

a_{9} = a + 8d = -8 → (2)

(2) – (1)

Put d = -2 in (1) ⇒ a + 2d = 4

⇒ a + 2(-2) = 4

⇒ a – 4 = 4

⇒ a = 4 + 4 = 8

Given n^{th} term a = a + (n – 1) d = 0

⇒ 8 + (n – 1)(-2) = 0

⇒ -2n + 2 = 8

⇒ -2n = -8 – 2

∴ n = \(\frac{-10}{-2}\) = 5

Therefore, 5^{th} term become zero in A.P.

Question 10.

The 17^{th} term of an AP exceeds its 10^{th} term by 7. Find the common difference.

Solution:

Given 17^{th} term of an A.P = 10^{th} term + 7

a + 16d = a + 9d = 7

a + 16d – a – 9d = 7

7d = 7

∴ d = \(\frac{7}{7}\) = 1

Therefore, common difference d = 1.

Question 11.

Which term of the AP: 3, 15, 27, 39,…… will be 132 more than its 54^{th} term?

Solution:

Given A.P is : 3, 15, 27, 39,…….

and n^{th} term = 54^{th} term +132

a + (n – 1)d = a + 53d + 132

a = 3, d = a_{2} – a_{1} = 15 – 3 = 12

⇒ 3 + (n – 1)12 = 3 + 53(12) + 132

⇒ (n- 1) 12 = 636 + 132

⇒ (n – 1) 12 = 768

⇒ n – 1 = \(\frac{768}{12}\) = 64

∴ n = 64 + 1 = 65

Therefore 65^{th} term is 132 more than its 54^{th} term.

Question 12.

Two APs have the same common dif-ference. The difference between their 100th terms is 100, what is the differ-ence between their 1000th terms?

Solution:

Let two A.Ps are a_{1}, a_{2},…… a_{n} and b_{1}, b_{2},……, b_{n}.

n^{th} term a_{n} = a + (n – 1) d

Difference of 100^{th} terms is 100

a_{n} – b_{n} = 100

a_{100} – b_{100} = 100

⇒ a_{1} + (100 – 1)d – [b_{1} + (100 – 1)d] = 100

⇒ a_{1} + 99d – b_{1} – 99d = 100

⇒ a_{1} – b_{1} = 100

difference of 1000^{th} terms

Therefore a_{1000} – b_{1000} = a_{1} – b_{1} = 100

Difference of 1000^{th} terms is 100.

Question 13.

How many three-digit numbers are divisible by 7 ?

Solution:

Three-digit numbers which are divisible by 7 are 105, 112, 119, ……….., 994.

a = 105, d = a_{2} – a_{1}

= 112 – 105 = 7

n^{th} term a_{n} = a + (n – 1) d

⇒ 105 + (n – 1) 7 = 994

⇒ (n – 1) 7 = 994 – 105

⇒ (n – 1) = \(\frac{889}{7}\) = 127

⇒ n = 127 + 1

∴ n = 128

Therefore, three-digit numbers divisible by 7 are 128.

Alternate method :

[=\frac{\text { Last number }- \text { First number }}{7}]

= \(\frac{999-100}{7}\) = 128.4

128 number divisible by 7.

Question 14.

How many multiples of 4 lie between 10 and 250?

Solution:

4 multiples between 10 and 250 are 12, 16, 20,…..248

a = 12, d = a_{2} – a_{1} = 16 – 12 = 4, l = 248

last term l = a + (n – 1)d

⇒ 12 + (n – 1)4 = 248

⇒ (n – 1)4 = 248 – 12

⇒ (n – 1)4 = 236

⇒ n – 1 = \(\frac{236}{4}\) = 59

∴ n = 59 + 1 = 60

∴ There are 60 numbers of 4 multiples between 10 and 250.

Question 15.

For what value of n, are the th terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?

Solution:

Given A.Ps are 63, 65, 67, . …… and 3, 10, 17,…

a = 63, d = 65 – 63 = 2

and b = 3, d = 10 – 3 = 7

and n^{th} terms are equal.

That is a_{n} = b_{n}

a + (n – 1) d = b + (n – 1)d

63 + (n – 1)2 = 3 + (n – 1)7

63 + 2n – 2 = 3 + 7n – 7

2n – 7n = -4 – 61

-5n = -65 ⇒ n = \(\frac{-65}{-5}\) = 13

Therefore, n = 13

Question 16.

Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Solution:

Given 3rd term in AP is 16.

a_{3} = a + 2d = 16 → (1)

7^{th} term exceeds the fifth term by 12

i.e., a_{7} = a_{5} + 12

a + 6d = a + 4d + 12

⇒ a + 6d – a – 4d = 12

⇒ 2d = 12 ⇒ d = \(\frac{12}{2}\) = 6

Put d = 6 in (1) ⇒ a + 2d = 16

⇒ a + 2(6) = 16

⇒ a = 16 – 12 = 4

A.P : a, a + d, a + 2d, a + 3d,. ..

4, 4 + 6, 4 + 2(6), 4 + 3(6),…….

Therefore, A.P is 4, 10, 16, 22,…….

Question 17.

Find the 20th term from the last term of the AP : 3, 8, 13,…, 253.

Solution:

Given A.P is : 3, 8, 13,… 253,

A.P from the last is : 253, 248, 243,…

a = 253, d = 248 – 253 = -5, n = 20

a_{n} = a + (n – 1) d

a_{20} = a + (20 – 1) d = a + 19d

= 253 + 19(-5)

= 253 – 95 = 158

Therefore, 20^{th} term from the last is 158.

Question 18.

The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Solution:

Let the A.P is a, a + d, a + 2d,………

Sum of 4^{th} and 8^{th} terms = 24

a_{4} + a_{8} = 24

a + 3d + a + 7d = 24

2a + 10d = 24

2(a + 5d) = 24

a + 5d = 12 → (1)

Sum of 6^{th} and 10^{th} terms = 44

a_{6} + a_{10} = 44

a + 5d + a + 9d = 44

2a + 14d = 44

2(a + 7d) = 44

a + 7d = 22 → (2)

(2) – (1) ⇒

Put, d = 5 in (1)

⇒ a + 5(5) = 12

⇒ a = 12 – 25 = -13

A.P : a, a + d, a + 2d,…

-13, -13 + 5, -13 + 2(5),………

Therefore, A.P is -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:

Subbarao’s annual salary (a) = ₹ 5000

increment (d) = ₹ 200

last year salary (l) = ₹ 7000

So, A.P is 5000, 5200, 5400,…….. 7000

Last term l = a + (n – 1) d

⇒ 5000 + (n – 1) 200 = 7000

⇒ (n- 1) 200 = 7000 – 5000

⇒ n – 1 = \(\frac{2000}{200}\) = 10

∴ n = 10 + 1 = 11

∴ Subbarao’s income reach ₹ 7000 in 11^{th} year.

Question 20.

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

Solution:

Ramkali first week savings (a) = ₹ 5

saving increased (d) = ₹ 1.75

weekly savings in the nth week (l) = ₹ 20.75

last term l = a + (n – 1) d

5 + (n – 1)(1.75) = 20.75

(n- 1) (1.75) = 20.75 – 5

(n – 1) (1.75) = 15.75

n – 1 = \(\frac{15.75}{1.75}\) = 9

n = 9 + 1 = 10

Therefore n = 10