RBSE Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.2

Question 1.
Fill in the blanks in the following table, given that a is the first term, d the common difference and a_n the nth term of the AP:

	a	d	n	an
(i)	7	3	8	…
(ii)	-18	…	10	0
(iii)	…	-3	18	-5
(iv)	-18.9	2.5	…	3.6
(v)	3.5	0	105	…

Solution:
(i) a = 7, d = 3, n = 8
∵ a_n = a + (n – 1) d
∴ a₈ = 7 + (8 – 1)3
= 7 + 21 = 28

(ii) a = – 18, n = 10, an = 0
∵ a_n = a + (n – 1) d
∴ a₁₀ = -18 + (10 – 1)d
or 0 = – 18 + 9d
or 9d= 18
or d = \(\frac{18}{9}\) = 2

(iii) d = -3, n = 18, a_n = – 5
∵ a_n = a + (n – 1) d
∴ a₁₈ = a + (18 – 1) (-3)
or -5 = a – 51
or ₂ = -5 + 51 = 46

(iv) a = -18.9, d = 2.5, a_n = 3.6
∵ a_n = a + (n – 1 )d
∴ 3.6 = -18.9 + (n – 1)2.5
or 3.6 + 18.9 = (n – 1)2.5
or (n – 1)2.5 =22.5
or n – 1 = \(\frac{22.5}{2.5}\) = 9
or n = 9 + 1 = 10

(v) a = 3.5, d= 0, n = 105
∵ a_n = a + (n – 1)d
∴ a_n = 3.5 + (105 – 1)0
a_n = 3.5 + 0 = 3.5

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
(ii) 11th term of the AP: – 3, –\(\frac{1}{2}\), 2, …….., is
(A) 28
(B) 22
(C) -38
(D) -48\(\frac{1}{2}\)
Solution:
(i) The given AP is 10, 7, 4,
Here, a₁ = 10, a₂ = 7, a₃ = 4
a₂ – a₁ = 7 – 10 = – 3
a₃ – a₂ = 4 – 7 = – 3
∵ a₂ – a₁ = a₃ – a₂ = – 3 = d (say)
∴ a_n = a + (n – 1) d
Now a₃₀ = 10 + (30 – 1) (- 3)
= 10 – 87 = -77
∴ Correct Ans. is (C).

Solution: (ii) The given AP is – 3, –\(\frac{1}{2}\), 2,….
Here, a₁ = -3, a₂ = – y, a₃ = 2,
a₂ – a₁ = –\(\frac{1}{2}\) + 3 = \(\frac{-1+6}{2}\) = \(\frac{5}{2}\)
a₃ – a₂ = 2 + \(\frac{1}{2}\) = \(\frac{4+1}{2}\) – \(\frac{5}{2}\)
∵ a₂ – a₁ = a₃ – a₂ = \(\frac{5}{2}\) = d (Say)
∵ a_n = a + (n – 1) d
Now a₁₁ = -3 + (11 – 1)\(\frac{5}{2}\)
= -3 + 10 × \(\frac{5}{2}\) = -3 + 25 = 22
∴ Correct Ans. is (B).

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

Solution:
Let the first term and the common difference of the given AP be a and d respectively.
(i) Here, a₁ = a = 2
and a₃ = a + 2d = 26
or 2 + 2d = 26
or 2d = 26 – 2 = 24
or d = \(\frac{24}{2}\) = 12
∴ Missing term a₂ = a + d = 2 + 12 = 14
Hence the missing term in the box a₂ = 14

(ii)Here, a₂ = a + d = 13 ….(1)
and a₄ = a + 3d = 3 …(2)
Now, from (2) – (1)
a + 3d = 3
a + d = 13
2d = – 10
⇒ d = \(\frac{-10}{2}\) = -5
Substituting this value of d in (1)
a – 5 = 13
a = 13 + 5= 18
∴ a₁ = a = 18
a₃ = a + 2d = 18 + 2 (-5)
= 18 – 10 = 8
Hence, the missing terms in the boxes are 18 and 8 respectively

(iii) Here, a₁ = a = 5

Hence, the missing terms in the boxes are 6\(\frac{1}{2}\) and 8

(iv) Here, a₁ = a = – 4
and a₆ = a + 5d = 6
or -4 + 5d = 6
or 5d = 6 + 4
or 5d = 10
or d = \(\frac{10}{5}\) = 2
Now, a₂ = a + d = – 4 + 2 = -2
a₃ = a + 2d = -4 + 2 (2)
= -4 + 4 = 0
a₄ = a + 3d = -4 + 3(2)
= -4 + 6 = 2
a₅ = a + 4d = -4 + 4(2)
= -4 + 8 = 4
Hence, the missing terms in the boxes are – 2, 0 and 2.

(v) Here, a₂ = a + d= 38 ….(1)
and a₆ = a + 5d = – 22 …(2)
Now, from (2) – (1)

Substituting this value of d in (1)
a + (-15) =38
a = 38 + 15 = 53
a₁ = a = 53
a₃ = a + 2d = 53 + 2(-15)
= 53 – 30 = 23
a₄ = a + 3d = 53 + 3(-15)
= 53 – 45 = 8
a₅ = a + 4d = 53 + 4(-15)
= 53 – 60 = – 7
Hence, the missing terms in the boxes are 53, 23, 8 and – 7

Question 4.
Which term of the AP : 3, 8, 13, 18,….., is 78?
Solution:
The given AP is : 3, 8, 13, 18, ……
Here, a₁ = 3, a₂ = 8, a₃ = 13, a₄ = 18
a₂ – a₁ = 8 – 3 = 5
a₃ – a₂ = 13 – 8 = 5
∵ a₂ – a₁ = a₃ – a₂ = 5 = d (say)
Using the formula a_n = a + (n – 1) d
or 78 = 3 + (n – 1) 5
or 5 (n – 1) = 78 – 3
or n – 1 = \(\frac{75}{5}\) = 15
or n = 15 + 1 = 16
Hence, the 16th term of the given AP 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) The given AP is : 7, 13, 19, ….
a₁ = 7, a₂ = 13, a₃ = 19
a₂ – a₁ = 13 – 7 = 6
a₃ – a₂ = 19 – 13 = 6
∵ a₂ – a₁ = a₃ – a₂ = 6 = d (say)
∴ Using the formula, a_n = a + (n – 1) d
⇒ 205 = 7 + (n – 1)6
or (n – 1)6 = 205 – 7
or (n – 1) = \(\frac{198}{6}\)
or n – 1 = 33
or n = 33 + 1 = 34
Hence, 34th term of the AP is 205.

(ii) The given AP is 18, 15\(\frac{1}{2}\), 13, ……..

or n – 1 = 26
or n = 26 + 1 = 27
Hence, 27th term of the AP is – 47

Question 6.
Check whether – 150 is a term of the AP: 11, 8, 5, 2, …..
Solution:
The given AP is 11, 8, 5, 2, ….
a₁ = 11, a₂ = 8, a₃ = 5, a₄ = 2
a₂ – a₁ = 8 – 11 = -3
a₃ – a₂ = 5 – 8 = -3
a₄ – a₃ = 2 – 5 = -3
∵ a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = -3 = d (say)
Let – 150 be the n^th term of the given AP
a_n = -150
⇒ a + (n – 1) d = -150
or 11 + (n – 1) (-3) = -150
or (n – 1) (-3) = -150 – 11 = -161
or n – 1 = \(\frac{161}{3}\)
or n = \(\frac{161}{3}\) + 1 = \(\frac{161+3}{3}\)
or n = \(\frac{161}{3}\) = 4\(\frac{2}{3}\)
Which is not a natural number.
Hence, -150 is not a term of the given AP.

Question 7.
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution:
Let ‘a’ and ‘d’ respectively be the first term and the common difference of the given AP.
Given that a₁₁ = 38
⇒ a + (11 – 1)d = 38
[∵ a_n = a + (n- 1) d]
⇒ a + 10d = 38 ….(1)
and a₁₆ = 73
a + (16 – 1)d = 73
[∵ a_n = a + (n – 1)d]
a + 15d = 73 ….(2)
Now, from (2) – (1)

Substituting the value of d in (1)
a + 10(7) =38
or a + 70 = 38
a = 38 – 70 = -32
Now a₃₁ = a + (31 – 1)d = -32 + 30(7)
= -32 + 210 = 178
Hence, the value of the 31st term of the AP will be 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:
Let ‘a’ and ‘d’ respectively be the first term and the common difference of the given AP.
Given that a₃ = 12
⇒ a + (3 – 1) = 12
[∵ a_n = a + (n – 1)d]
or a + 2d = 12 …..(1)
and last term = a₅₀ = 106
a + (50 – 1)d = 106
[∵ a_n = a + (n – 1) d]
or a + 49d = 106 ….(2)
Now, from (2) – (1)

Substituting this value of d in (1)
a + 2 (2) = 12
or a + 4 = 12
or a = 12 – 4 = 8
Now, a₂₉ = a + (29 – 1) d
= 8 + 28 (2)
= 8 + 56 = 64
Hence, 29th term of the AP will be 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:
Let ‘a’ and ‘d’ respectively be the first term and common difference of the given AP.
Given that a₃ = 4
⇒ a + (3 – 1)d =4
[∵ a_n = a + (n – 1)d]
⇒ a + 2d = 4 ….(1)
and a₉ = – 8
a + (9 – 1)d = -8
[∵ a_n = a + (n – 1)d]
or a + 8d = – 8 ….(2)
Now, from (2) – (1)

Substituting this value of d in (1)
a + 2 (-2) = 4
or a – 4 = 4
or a = 4 + 4 = 8
Now, a_n = 0 (Given)
a + (n – 1)d = 0
or 8 + (n – 1) (-2) = 0
or -2 (n – 1) = -8
or n – 1 = 4
or n = 4 + 1 = 5
Hence, 5th term of the AP is zero.

Question 10.
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution:
Let ‘a’ and ‘d’ respectively be the first term and common difference of the given AP.
Now, a₁₇ = a + (17 – 1) d
= a + 16 d
and a₁₀ = a + (10 – 1) d
= a + 9d
According to the question,
a₁₇ – a₁₀ = 7
(a + 16d) – (a + 9d) = 1
or a + 16d – a – 9d = 7
or 7d = 7
or d = \(\frac{7}{7}\) = 1
Hence, the common difference is 1.

Question 11.
Which term of the AP : 3, 15, 27, 39, will be 132 more than its 54th term?
Solution:
Let ‘a’ and ‘d’ respectively be the first term and common difference of the given AP.
The given AP is : 3, 15, 27, 39, …..
a₁ = 3, a₂ = 15, a₃ = 27, a₄ = 39
a₂ – a₁ = 15 – 3 = 12
a₃ – a₂ = 27 – 15 = 12
∴ d = a₂ – a₁ = a₃ – a₂ = 12
Now, a₅₄ = a + (54 – 1)d
= 3 + 53(12)
= 3 + 636 = 639
According to the question,
a_n = a₅₄ + 132
or a + (n – 1)d = 639 + 132
or 3 + (n – 1) (12) = 771
or (n – 1) 12 = 771 – 3 = 768
or n – 1 = \(\frac{768}{12}\) = 64
or n = 64 + 1-65
Hence, 65th term of the series is 132 more than the 54th term.

Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let ‘a’ and ‘d’ respectively be the first term and common difference of the first AP.
Also, let ‘a’ and d’ respectively be the first term and the common difference of the second AP.
[Since both APs have the same common difference]
According to the question
[a₁₀₀ of the second AP] – [a₁₀₀ of the first AP] = 100
or [A + (100 – 1)d] – [a + (100 – 1)d] = 100
or A + 99d – a – 99d =100
or A – a =100 …(1)
Now, [a₁₀₀₀ of the second AP] – [a₁₀₀₀ of the first AP]
= [A + 1000 – 1)d] – (a + (1000 – 1 )d]
= A + 999d – a – 999d
= A – a
= 100 [Using (1)]
Hence, difference of 1000th terms = 100

Question 13.
How many three-digit numbers are divisible by 7?
Solution:
Three digit numbers divisible by 7 are : 105, 112, 119, ….., 994
Here, a = a₁ = 105, a₂ = 112, a₃ = 119
and a_n = 994
a₂ – a₃ = 112 – 105 = 7
a₃ – a₂ = 119 – 112 = 7
∴ d = a₂ – a₁ = a₃ – a₂ = 1
Given that a_n = 994
⇒ a + (n – 1) d = 994
or 105 + (n – 1)7 = 994
or (n – 1)7 = 994 – 105
or (n – 1)7 = 889
or n – 1 = \(\frac{889}{7}\) = 127
or n = 127 + 1 = 128
Hence, there are 128 three digit numbers divisible by 7.

Question 14.
How many multiples of 4 lie between 10 and 250?
Solution:
First multiple of 4 greater than 10 = 12
First multiple of 4 less than 250 = 248
∴ The series of multiples of 4 between 10 and 250 will be the following :
12, 16, 20, 24,…., 248
Here a = a₁ = 12, a₂ = 16, a₃ = 20
and a_n = 248
a₂ – a₁ = 16 – 12 = 4
a₃ – a₂ = 20 – 16 = 4
∴ d = a₂ – a₁ = a₃ – a₂ = 4
Given that a_n = 248
⇒ a + (n – 1) d = 248
or 12 + (n – 1)4 =248
or 4 (n – 1) = 248 – 12 = 236
or n – 1 = \(\frac{236}{4}\) = 59
or n = 59 + 1 = 60
Hence, there are 60 multiples of 4 between 10 and 250.

Question 15.
For what value of n, are the nth terms of two APs: 63, 65, 67, ….. and 3, 10, 17, ….. equal?
Solution:
The given first AP is : 63, 65, 67, ….
Here, a = a₁ = 63, a₂ = 65, a₃ = 67
a₂ – a₁ = 65 – 63 = 2
a₃ – a₂ = 61 – 65 = 2
∴ d = a₂ – a₁ = a₃ – a₂ = 2
and second AP is : 3, 10, 17,….
Here a = a₁ = 3, a₂ = 10, a₃ = 17
a₂ – a₁ = 10 – 3 = 7
a₃ – a₂ = 17 – 10 = 7
∴ d = a₂ – a₁ = a₃ – a₂ = 7
According to the question,
[With term of first AP] = [With term of second AP]
⇒ 63 + (n – 1) 2 = 3 + (n – 1) 7
or 63 + 2n – 2 = 3 + 7n – 7
or 61 + 2n = 7n – 4
or 2n – 7n = – 4 – 61
or – 5n = – 65
or n = \(\frac{65}{5}\) = 13
Hence, the 13th term of both the APs will be equal.

Question 16.
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
Let ‘a’ and ‘d’ respectively be the first term and the common difference of the given AP.
Given that a₃ = 16
⇒ a + (3 – 1)d = 16
⇒ a + 2d = 16 ….(1)
According to the question
a₇ – a₅ = 12
[a + (7 – 1)d] – [a + (5 – 1)d] = 12
a + 6d – a – 4d = 12
2d = 12
d = \(\frac{12}{2}\) = 6
Substituting this value of d in (1)
a + 2 (6) = 16
⇒ a = 16 – 12 = 4
Hence, the given AP is :
4, 10, 16, 22, 28,…..

Question 17.
Find the 20th term from the last term of the AP: 3,8,13, …., 253.
Solution:
The given AP is : 3, 8, 13, ……. 253
Here, a = a₁ = 3, a₂ = 8, a₃ = 13
and a_n = 253
a₂ – a₁ = 8 – 3 = 5
a₃ – a₂ = 13 – 8 = 5
∵ d = a₂ – a₁ = a₃ – a₂ = 5
Now a_n = 253
⇒ 3 + (n – 1)5 = 253
⇒ (n – 1) 5 = 250
n – 1 = = 50
[∵ a_n = a + (n – 1)d]
⇒ n = 50 + 1 = 51
∴ 20th term from the last term of A.P
= (Total number of terms) – 20 + 1
= 51 – 20+ 1 = 32^th term
∴ 20th term from the last term of A .P
= 32th term from the beginning
= 3 + (32 – 1)5
[∵ a_n = a + (n – 1) d]
= 3 + 31 x 5 = 3 + 155= 158
Hence, 20^th term from the last term of the AP= 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 ‘a’ and ‘d’ respectively be the first term and the common difference of the given AP.
According to the first condition of the problem
a₄ + a₈ = 24
a + (4 – 1)d + a + (8 – 1)d = 24
[∵ a_n = a + (n – 1)d]
or 2a + 3d + 7d = 24
or 2a + 10d = 24
or a + 5d = 12 ….(1)
According to the second condition of the problem,
a₆ + a₁₀ = 44
⇒ a + (6 – 1)d + a + (10 – 1)d = 44
[∵ a_n = a + (n – 1) d]
or 2a + 5d + 9d = 44
or 2a + 14d = 44
or a + 7d = 22 ….(2)
Now, from (2) – (1)

Substituting this value of d in (1)
a + 5 (5) = 12
a + 25 = 12
a = 12 – 25 = – 13
∴ a₁ = a = – 13
a₂ = a + d
= – 13 + 5 = -8
a₂ = a + 2d = – 13 + 2 (5)
= – 13 + 10 = – 3
Hence, the first three terms of the AP 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:
Let the original salary of Subba Rao = ₹ 5000
Annual increment = ₹ 200
Let ‘n’ denote the number of years.
∴ First term = a = ₹ 5000
Common difference = d = ₹ 200
and a_n = ₹ 7000
5000 + (n – 1)200 = 7000
[∵ a_n = a + (n – 1) d]
or (n – 1)200 = 7000 – 5000
or (n – 1)200 = 2000
or n – 1 = \(\frac{2000}{200}\) = 10
or n = 10+ 1 = 11
Hence, in the 11th year the salary of Subba Rao will be ₹ 7000.

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:
Savings in first week = ₹ 5
Increase in weekly savings = ₹ 1.75
It is obvious that it is an AP whose terms are n.
a₁ = 5 ,d = 1.75
∴ a₂ = 5 + 1.75 = 6.75
a₃ = 6.75 + 1.75 = 8.50
Also, a_n = 20.75 (Given)
5 + (n – 1) 1.75 = 20.75
[∵ a_n = a + (n – 1) d]
or (n – 1) 1.75 = 20.75 – 5
or (n – 1) 1.75 = 15.75
or (n – 1) = \(\frac{1575}{100}\) × \(\frac{100}{175}\)
or n – 1 = 9
or n = 9 + 1 = 10
Hence, in 10th week, Ramkali saves ₹ 20.75.