Chapter 5 Arithmetic Progressions NCERT Exemplar Solutions Exercise 5.1 Class 10 Maths
Chapter Name  NCERT Maths Exemplar Solutions for Chapter 5 Arithmetic Progressions Exercise 5.1 
Book Name  NCERT Exemplar for Class 10 Maths 
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Related Study  NCERT Solutions for Class 10 Maths 
Exercise 5.1 Solutions
Multiple Choice Questions
Choose the correct answer from the given four options in the following questions:
1. In an AP, if d = 4, n = 7, a_{n} = 4, then a is
A. 6
B. 7
C. 20
D. 28
Solution
(D) 28
In an A.P,
a_{n} = a + (n  1)d
(a = first term, an is nth term and d is the common difference)
4 = a + (7  1)(4)
⇒ 4 = a  24
⇒ a = 24 + 4 = 28
2. In an AP, if a = 3.5, d = 0, n = 101, then a_{n} will be
A. 0
B. 3.5
C. 103.5
D. 104.5
Solution
(B) 3.5
In an A.P,
a_{n} = a + (n  1)d
(a = first term, an is nth term and d is the common difference )
a_{n} = 3.5 + (101  1)0 = 3.5
(Since, d = 0, it's a constant A.P)
3. The list of numbers 10, 6, 2, 2, ..... is
A. an AP with d = 16
B. an AP with d = 4
C. an AP with d = 4
D. not an AP
Solution
In the given A.P,
a_{1} = 10
a_{2} = 6
a_{3} = 2
a_{4} = 2
a_{2}  a_{1} = 4
a_{3}  a_{2} = 4
a_{4}  a_{3} = 4
a_{2}  a_{1} = a_{3}  a_{2} = a_{4}  a_{3} = 4
So, it is an A.P with d = 4
4. The 11th term of he A.P. : 5, 5/2, 0, 5/2 , ....... is
A. 20
B. 20
C. 30
D. 30
Solution
According to the given A.P.
a =  5
d = 5  (5/2) = 5/2
n = 11
Also,
a_{n} = a + (n  1)d
Here, (a = first term, a_{n} is nth term and d is the common difference)
a_{11} = 5+ (11  1)(5/2)
⇒ a_{11} = 5 + 25 = 20
5. The first four terms of an AP, whose first term is –2 and the common difference is –2, are
A. –2, 0, 2, 4
B. –2, 4, –8, 16
C. –2, –4, –6, –8
D. –2, –4, –8, –16
Solution
First term,
a =  2
Second term,
d = 2
a_{1} = a = 2
Also,
a_{n} = a + (n  1)d
where,
a = first term, a_{n} is nth term, d is the common difference
Therefore,
a_{2} = a + d
= 2 + (2) =  4
Similarly,
a_{3} = 6
a_{4} = 8
So, the A.P is 2, 4, 6, 8.
6. The 21st term of the AP whose first two terms are 3 and 4 is
A. 17
B. 137
C. 143
D. 143
Solution
First two terms of an AP are a = 3 and a_{2} = 4.
We know, nth term of an AP is
a_{n} = a + (n  1)d
Here, a = first term, a_{n} is nth term, d is the common difference
a_{2} = a + d
⇒ 4 = 3 + d
⇒ d = 7
Common difference,
d = 7
a_{21 } = a + 20d
= 3 + 20 × 7
= 137
7. If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term ?
A. 30
B. 33
C. 37
D. 38
Solution
In an A.P.
a_{n} = a + (n  1)d
Here, a = first term, a_{n} is nth term, d is he common difference
a_{2} = a + d = 13 ...(i)
a_{5} = a + 4d = 25 ...(ii)
From equation (i),
a = 13  d
Using this in equation (ii),
13 – d + 4d = 25
⇒ 13 + 3d = 25
⇒ 3d = 12
⇒ d = 4
a = 13 – 4 = 9
a_{7} = a + 6d
= 9 + 6(4)
= 9 + 24 = 33
8. Which term of the AP. 21, 42, 63, 84.... is 210 ?
A. 9th
B. 10th
C. 11th
D 12th
Solution
Let nth term of the given AP be 210.
According to question,
First term,
a = 21
Common difference,
d = 42 – 21 = 21
a_{n} = 210
We know that the nth term of an AP is a_{n} = a + (n – 1)d
Where, a = first term, an is nth term, d is the common difference
210 = 21 + (n – 1)21
⇒ 189 = (n – 1)21
⇒ n – 1 = 9
⇒ n = 10
So, 10th term of an AP is 210.
9. If the common difference of an AP is 5, then what is a_{18}  a_{13} ?
A. 5
B. 20
C. 25
D. 30
Solution
Given, d = 5
Now,
As we know, nth term of an AP is
a_{n} = a + (n – 1)d
Here, a = first term, an is nth term, d is the common difference
a_{18} a_{13} = a + 17d – (a + 12d)
= 5d
= 5(5)
= 25
10. What is the common difference of an AP in which a_{18}  a_{14} = 32 ?
A. 8
B. 8
C. 4
D. 4
Solution
(a) a_{18}  a_{14} = 32
[a + (18 – 1)d] – [a + (14 – 1)d] = 32 [a_{n} = a + (n – 1)d]
a + 17d – a – 13d = 32
⇒ 4d = 32
⇒ d = 8
So, (a) is the correct answer.
11. Two APs have the same common difference. The first term of one of these is –1 and that of the other is – 8. Then the difference between their 4th terms is
A. 1
B. 8
C. 7
D. 9
Solution
(C)
According to question,
a_{1} (for the first AP) = 1 and a_{1} (for the second AP) = 8
Let d be the same common difference of two A.Ps.
d_{1} = d,
a_{n} = a + (n  1)d
Now,
a_{4} of first AP  a_{4} of second AP
(1 + 3d)  (8 + 3d)
⇒ 1 + 3d + 8 3d = 7
So, the required answer is (c).
12. If 7 times the 7th term of an AP is equal to 11 times its 11 th term, then its 18th term will be
A. 7
B. 11
C. 18
D. 0
Solution
(d)
a_{18} = a + (18 – 1)d
= a + 17d
Also,
7a_{7} = 11a_{11} (Given)
⇒ 7[a + (7 – 1)d] = 11[a + (11 – 1)d]
⇒ 7[a + 6d] = 11[a + 10d]
⇒ 7a + 42d = 11a + 110d
⇒ 0 = 11a – 7a + 110d – 42d
⇒ 0 = 4a + 68d
⇒ 0 = a + 17d
⇒ a_{18} = 0
So, (d) is the correct answer.
13. The 4th term from the end of the AP: –11, –8, –5, ..., 49 is
A. 37
B. 40
C. 43
D. 58
Solution
Reversing the A.P., we get
49 ......, 5, 8, and 11
d = 8  (5)
= 8 + 5 = 3
a = 49 and n = 4
a_{n} = a + (n  1)d
⇒ a_{4} = 49 + (4  1)(3)
⇒ a_{4} = 49 + 3(3)
= 49  9
⇒ a_{4} = 40
So, the required value of a_{4} is 40 and answer is (b)
14. The famous mathematician associated with finding the sum of the first 100 natural numbers is
A. Pythagoras
B. Newton
C. Gauss
D. Euclid
Solution
(C)
Gauss is the famous mathematician associated with finding the sum of first 100 natural numbers, i.e., 1 + 2 + 3 + 4 + 5 + ... + 100
a = 1, d = 1, n = 100
= 50 × 101
= 5050
15. If the first term of an AP is –5 and the common difference is 2, then the sum ofthe first 6 terms is
A. 0
B. 5
C. 6
D. 15
Solution
(a)
a = 5,
d = 2
n = 6
We have,
= 3[10 + 5 × 2]
= 3[ 10 + 10]
= 3[0]
S_{6} = 0
So, (a) is the correct answer.
16. The sum of first 16 terms of the AP: 10, 6, 2,... is
A. –320
B. 320
C. –352
D. –400
Solution
(a)
a = 10,
n = 16,
d = 6 – 10 = –4
So, the required answer is (a)
17. If AP if a = 1 a_{n} = 20 and S_{n} = 399, then n is
A. 19
B. 21
C. 38.
D. 42
Solution
(c)
S_{n} = [2a + (n  1)d]
S_{n} = [a + a + (n  1)d]
⇒ 399 = [a + a_{n} ] [a_{n} = last term]
⇒ 399 = [1 + 20]
⇒ n = 38
So, (c) is the correct answer.
18. The sum of first five multiples of 3 is
A. 45
B. 55
C. 65
D. 75
Solution
(a)
1st five multiples of 3 are 3, 6, 9, 12, 15, .....
a = 3,
n = 5,
d = 6 – 3 = 3
= 45
(a) is the correct answer.