# AP 10th Class Maths 5th Chapter Arithmetic Progressions Exercise 5.2 Solutions

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 an is the nth 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 = a2 – a1 = 7 – 10 = -3
an = a + (n – 1)d
a30 = 10 + (30 – 1)(-3) = 10 + 29 × (-3)
a30 = 10 – 87 = -77

ii) 11th 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 = a2 – a1
= –$$\frac{1}{2}$$ – (-3) = –$$\frac{1}{2}$$ + 3 = $$\frac{5}{2}$$
an = a + (n – 1)d
a11 = -3 + (11 – 1)$$\frac{5}{2}$$

a11 = -3 + 25 = 22

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

Solution:
i) Given 2, ____, 26
a1 = 2, a3 = 26
d = a2 – a1 = a3 – a2
____ – 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, a2, a3, a4, a5, 6
a1 = a = -4, a + 5d = 6
-4 + 5d = 6
5d = 6 + 4 = 10
d = $$\frac{10}{5}$$ = 2
a2 = a1 + d = -4 + 2 = -2
a3 = a1 + 2d = -4 + 2(+2) = -4 + 4 = 0
a4 = a1 + 3d = -4 + 3(2) = -4 + 6 = 2
a5 = a1 + 4d = -4 + 4(2) = -4 + 8 = 4
∴ A.P is -4, -2, 0, 2, 4, 6.

v) Given a1, 38, a3, a4, a5, -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 = a2 – a1 = 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
∴ 16th 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 = a2 – a1 = 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 = a2 – a1 = 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 a11 = 38 and a16 = 73
a11 = a + 10d = 38
a16 = a + 15d = 73
(2) – (1) ⇒

Put d = -7 in (1) ⇒ a11 = a + 10d = 38
⇒ a + 10(7) = 38
⇒ a = 38 – 70 = -32
31st term is a31 = a + 30d = -32 + 30(7)
a31 = -32 + 210 = 178
Therefore, 31st 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 3rd term of A.P: a3 = 12,
last term l = 106
Number of terms n = 50
a3 = 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
29th term is a29 = a + 28d
= 8 + 28(2)
= 856 = 64
Therefore, 29th 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 3rd term of an A.P is a3 = 4
and 9th term is a9 = -8
a3 = a + 2d = 4 → (1)
a9 = 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 nth term a = a + (n – 1) d = 0
⇒ 8 + (n – 1)(-2) = 0
⇒ -2n + 2 = 8
⇒ -2n = -8 – 2
∴ n = $$\frac{-10}{-2}$$ = 5
Therefore, 5th term become zero in A.P.

Question 10.
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution:
Given 17th term of an A.P = 10th 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 54th term?
Solution:
Given A.P is : 3, 15, 27, 39,…….
and nth term = 54th term +132
a + (n – 1)d = a + 53d + 132
a = 3, d = a2 – a1 = 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 65th term is 132 more than its 54th 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 a1, a2,…… an and b1, b2,……, bn.
nth term an = a + (n – 1) d
Difference of 100th terms is 100
an – bn = 100
a100 – b100 = 100
⇒ a1 + (100 – 1)d – [b1 + (100 – 1)d] = 100
⇒ a1 + 99d – b1 – 99d = 100
⇒ a1 – b1 = 100
difference of 1000th terms

Therefore a1000 – b1000 = a1 – b1 = 100
Difference of 1000th 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 = a2 – a1
= 112 – 105 = 7
nth term an = 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 = a2 – a1 = 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 nth terms are equal.
That is an = bn
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.
a3 = a + 2d = 16 → (1)
7th term exceeds the fifth term by 12
i.e., a7 = a5 + 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
an = a + (n – 1) d
a20 = a + (20 – 1) d = a + 19d
= 253 + 19(-5)
= 253 – 95 = 158
Therefore, 20th 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 4th and 8th terms = 24
a4 + a8 = 24
a + 3d + a + 7d = 24
2a + 10d = 24
2(a + 5d) = 24
a + 5d = 12 → (1)
Sum of 6th and 10th terms = 44
a6 + a10 = 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 11th 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