NCERT Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2

In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.2. These solutions are based on new NCERT Syllabus.

NCERT Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 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 n^th term of the A.P.
Solution:

Solution:
(i) a_n = a + (n- 1)d
a₈ = 7 + (8 – 1)3 = 7 + 7 × 3 = 7 + 21
⇒ a₈ = 28

(ii) a_n = a + (n – 1)d
⇒ a₁₀ = -18 + (10 – 1)7 ⇒ 0 = -18 + 9d
⇒ 9d = 18 ⇒ d = \(\frac{18}{9}=2\)
∴ d = 2

(iii) a_n = a + (n – 1)d
⇒ -5 = a + (18 – 1) × (-3)
⇒ -5 = a + 17 × (-3)
⇒ -5 = a – 51 ⇒ a = -5 + 51 = 46
Thus, a = 46

(iv) a_n = a + (n – 1)d
⇒ 3.6 = -18.9 + (n – 1) × 2.5
⇒ (n – 1) × 2.5 = 3.6 + 18.9

⇒ n = 9 + 1 = 10
Thus, n = 10

(v) a_n = a + (n- 1)d
⇒ a_n = 3.5 + (105 – 1) × 0
⇒ a_n = 3.5 + 104 × 0 ⇒ a_n = 3.5 + 0 = 3.5
Thus, a_n = 3.5

Question 2.
Choose the correct choice in the following and justify:
(i) 30^th term of the AP: 10,7,4, , is, ….,
(A) 97
(B) 77
(C) -77
(D) -87

(ii) 11^th term of the AP: -3, \(-\frac{1}{2}\), 2, …. ,is
(A) 28
(B) 22
(C) -38
(D) -48\(\frac{1}{2}\)
Solution:
(i) (C): Here, a = 10, n = 30
∵ T₁₀ = a + (n – 1)d and d = 7 – 10 = -3
∴ T₃₀ = 10 + (30 – 1) × (-3)
⇒ T₃₀ = 10 + 29 × (-3)
⇒ T₃₀ = 10 – 87 = -77

(ii) (B): Here, a = -3, n = 11 and

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

Solution:
(i) Here, a = 2, T₃ = 26
Let common difference = d
∴ T_n = a + (n- 1 )d
⇒ T₃ = 2 + (3 – 1)d
⇒ 26 = 2 + 2 d
⇒ 2d = 26 – 2 = 24

(ii) Let the first term = a
and common difference = d
Here, T₂ = 13 and T₄ = 3
T₂ = a + d = 13, T₄ = a + 3d = 3
T₁ – T₂ = (a + 3d) – (a + d) = 3 – 13

(iii)

(iv) Here, a = – 4, T₆ = 6
∵ T_n = a + (n -1 )d
T₆ = – 4 + (6 – 1)d ⇒ 6 = -4 + 5d ⇒ 5d = 6 + 4 = d = 10 – 5 = 2
T₂ = a + d = -4 + 2 =-2
T₃ = a + 2d = -4 + 2(2) = 0
T₄ = a + 3d = -4 + 3(2) = 2
T₅ = a + 4d = -4 + 4(2) = 4

(v) Here, T₂ = 38 and T₆ = -22
∴ T₂ = a + d = 38, T₆ = a + 5d = -22
⇒ T₆ – T₂ = a + 5d – (a + d) = -22 – 38 -60
⇒ 4d = -60 ⇒ d = \(\frac{-60}{4}\) = -15
a + d = 38 ⇒ a + (-15) = 38
⇒ a = 38 + 15 = 53
Now,
T₃ = a + 2d = 53 + 2(-15) = 53 – 30 = 23
T₄ = a + 3d = 53 + 3(-15) = 53 – 45 = 8
T₅ = a + 4d = 53 + 4(-15) = 53 – 60 = -7
Thus missing terms are

Question 4.
Which term of the AP: 3, 8, 13, 18, is 78?
Solution:
Let the n^th term be 78
Here, a = 3 ⇒ T₁ = 3 and T₂ = 8
∴ d = T₂ – T₁ = 8 – 3 = 5
And, T_n = a + (n- 1 )d
⇒ 78 = 3 + (n – 1) × 5 ⇒ 78 – 3 = (n -1) × 5
⇒ 75 = (n – 1) × 5 ⇒ (n – 1) = 75 ÷ 5 = 15
⇒ n = 15 + 1 = 16
Thus, 78 is the 16^th term of the given AP.

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) Here, a = 7,d = 13 – 7 = 6
Let total number of terms be n.
∴ T_n = 205
Now, T_n = a + (n – 1) ×d
= 7 + (n – 1) × 6 = 205
⇒ (n – 1) × 6 = 205 – 7 = 198
⇒ n – 1 = \(\frac{198}{6}=33\)
∴ n = 33 + 1 = 34
Thus, the required number of terms is 34.

(ii)

Thus, the required number of terms is 27.

Question 6.
Check whether -150 is a term of the AP:
11, 8, 5, 2…
Solution:
For the given AP,
we have a = 11, d = 8 -11 = -3
Let -150 be the n^th term of the given AP
∴ T_n = a + (n – 1 )d
⇒ -150 = 11 + (n – 1) × (-3)
⇒ -150 – 11 = (n – 1) × (-3)
⇒ -161 = (n – 1) ⇒ (-3)

But n should be a positive integer.
Thus, -150 is not a term of the given AP

Question 7.
Find the 31^st term of an AP whose 11^th term is 38 and the 16^th term is 73.
Solution:
Here, T₁₁ = 38 and T₁₆ = 73
Let the first term = a and the common difference = d.
T_n = a + (n – 1 )d
Then, T_n = a + (11 – 1)d = 38
⇒ a + 10d = 38 …(1)
and T₁₆ = a + (16 – 1)d = 73
⇒ a + 15d = 73 …(2)
Subtracting (1) from (2), we get
(a + 15d) – (a + 10d) = 73 – 38

Question 8.
An AP consists of 50 terms of which 3^rd term is 12 and the last term is 106. Find the 29^th term.
Solution:
Here, n = 50, T₃ = 12, T_n = 106
⇒ T₅₀ = 106
Let the first term = a and the common difference = d
T_n = a + (n – 1 )d
T₃ = a + 2d = 12 …(1)
T₅₀ = a + 49d = 106 …(2)
Subtracting (1) from (2), we get

Question 9.
If the 3^rd and the 9^th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
Here, T₃ = 4 and T₉ = -8
T_n = a + (n – 1)d
T₃ = a + 2d = 4 …. (1)
T₉ = a + 8d = – 8 …. (2)
Subtracting (1) from (2), we get

Question 10.
The 17^th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution:
Let a be the first term and d the common difference of the given AP
Now, using _n = a + (n – 1 )d, we have

Thus, the common difference is 1.

Question 11.
Which term of the AP : 3, 15, 27, 39,… will be 132 more than its 54^th term?
Solution:
Here, a = 3, d = 15 – 3 = 12
Using T_n = a + (n – 1 )d, we get

Thus, 132 more than 54^th term is the 65^th 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 for the 1^st AP, the first term = a and common difference = d

And for the 2^nd AP, the first term = a and common difference = d

According to the condition,

Question 13.
How many three-digit numbers are divisible by 7?
Solution:
The first three-digit number divisible by 7 is 105.
The last such three-digit number divisible by 7 is 994.
∴ The AP is 105,112,119, ,994
Let n be the required number of terms Here, a = 105, d = 7 and _n = 994

Thus, 128 numbers of 3-digits are divisible by 7.

Question 14.
How many multiples of 4 lie between 10 and 250?
Solution:
The first multiple of 4 beyond 10 is 12.
The multiple of 4 just below 250 is 248.

Thus, the required number of terms is 60.

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

Thus, the 13^th terms of the two given AP’s are equal.

Question 16.
Determine the AP whose third term is 16 and the 7^th term exceeds the 5^th term by 12.
Solution:
Let the first term = a and the common difference = d

Question 17.
Find the 20^th term from the last term of the AP : 3, 8, 13, …, 253.
Solution:
We have, the last term (l) = 253
Here, d = 8 – 3 = 5
Since, the n^th term before the last term is given by l – (n – 1 )d
We have 20^th term from the last term = l – (20 – 1) × 5 = 253 – 19 × 5 = 253 – 95 = 158

Question 18.
The sum of the 4^th and 8^th 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 first term = a and the common difference = d

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:
Here, a = ₹ 5000 and d = ₹ 200
Let, in the nth year Subba Rao gets ₹ 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 n^th week, her weekly savings become ₹ 20.75, find n.
Solution:
Here, a = ₹ 5 and d = ₹ 1.75

Thus, the required number of weeks is 10.