Nth Term of Arithmetic Progression - Class 10 Maths MCQ

This set of Class 10 Maths Chapter 5 Multiple Choice Questions & Answers (MCQs) focuses on “Nth Term of Arithmetic Progression”.

1. What will be the n^th term of the AP 1, 5, 9, 13, 17…….?
a) 4n – 3
b) 3n – 4
c) 4n + 3
d) 3n + 4
View Answer

Answer: a
Explanation: Here a = 1 and d = 4
The n^th term of the AP = a + (n – 1)d = 1 + (n – 1)4 = 1 + 4n – 4 = 4n – 3

2. If the n^th term of the AP is 2n + 7, then the common difference will be _____
a) -2
b) 2
c) 3
d) -3
View Answer

Answer: b
Explanation: n^th term of the AP is 2n + 7
T_n = 2n + 7
Common difference = T₂ – T₁ = (2 × 2 + 7 – (2 × 1 + 7)) = 11 – 9 = 2

3. If the n^th term of the AP is 7n – 9, then the first term is ________
a) 1
b) 6
c) 3
d) 4
View Answer

Answer: b
Explanation: n^th term of the AP is 7n – 9
T_n = 7n – 1
T₁ = 7(1) – 1 = 6
The first term of AP is 6.

4. If the n^th term of the AP is 8n + 1, then the 20^th term will be ______
a) 160
b) 120
c) 161
d) 121
View Answer

Answer: c
Explanation: n^th term of the AP is 8n + 1
T_n = 8n + 1
T₂₀ = 8(20) + 1
T₂₀ = 161

5. What will be the 99^th term from the end of the AP 500, 489, 478, 467… – 1139?
a) 1078
b) 1123
c) 12
d) 61
View Answer

Answer: d
Explanation: In this case since we have to find the 99^th term from the end.
We will consider the first term to be -1139 and the common difference will be 11
Now, a = -1139, d = 11 and n = 99
T₉₉ = a + (n – 1)d
T₉₉ = -1139 + (99 – 1)11
T₉₉ = -1139 + 1078 = -61
The value of 99^th term from the end is 61.

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6. If the p^th term of an AP is q and its q^th term is p, then what will be the value of its (p + q)^th term?
a) 1
b) p + q – 1
c) 0
d) 2(p + q – 1)
View Answer

Answer: c
Explanation: p^th term = q
a + (p – 1)d = q
a + pd – d = q (1)
q^th term = p
a + (q – 1)d = p
a + qd – d = p (2)
Subtracting (2) from (1) we get,
a + qd – d – (a + pd – d) = p – q
qd – pd = p – q
d = -1
Substituting in equation 1, we get,
a = p + q – 1
(p + q)^th term = a + (n – 1)d = p + q – 1 + (p + q – 1)(-1) = p + q – 1 – p – q + 1 = 0

7. If 5 times the 5^th term of an AP is equal to 15 times its 15^th term, then the value of its 20^th term will be _______
a) 0
b) 1
c) 2
d) 3
View Answer

Answer: a
Explanation: 5(5^th term) = 15(15^th term)
5^th term = a + (5 – 1)d = a + 4d
15^th term = a + (15 – 1)d = a + 14d
5(a + 4d) = 15(a + 14d)
5a + 20d = 15a + 210d
20d – 210d = 15a – 5a
-190d = 10a
a = -19d
Now, the 20^th term = a + (20 – 1)d = a + 19d
But a = -19d
Hence, -19d + 19d = 0

8. If the 11^th term of an AP is \(\frac {1}{13}\) and its 13^th term is \(\frac {1}{11}\), then what will be the value of 143^th term?
a) \(\frac {1}{143}\)
b) 1
c) 0
d) \(\frac {23}{143}\)
View Answer

Answer: b
Explanation: Here 11^th term = \(\frac {1}{13}\)
13^thterm = \(\frac {1}{11}\)
Let the first term of the AP be a and common difference be d
T₁₁ = a + (n – 1)d = \(\frac {1}{13}\)
T₁₁ = a + (11 – 1)d = \(\frac {1}{13}\)
T₁₁ = a + 10d = \(\frac {1}{13}\) (1)
T₁₃ = a + (n – 1)d = \(\frac {1}{11}\)
T₁₃ = a + (13 – 1)d = \(\frac {1}{11}\)
T₁₃ = a + 12d = \(\frac {1}{11}\) (2)
Subtracting (1) from (2)
We get,
a + 12d – (a + 10d) = \(\frac {1}{11} – \frac {1}{13}\)
2d = \(\frac {2}{143}\)
d = \(\frac {1}{143}\)
Now, substituting value of d in equation 1
We get,
T₁₁ = a + 10(\(\frac {1}{143}\)) = \(\frac {1}{13}\)
a = \(\frac {1}{143}\)
The 143^th term = \(\frac {1}{143}\) + 142\(\frac {1}{143} = \frac {(1+142)}{143} \) = 1

9. If the 7^th term of an AP is 20 and its 11^th term is 40 then, what will be the common difference?
a) 3
b) 4
c) 5
d) 2
View Answer

Answer: c
Explanation: Here 7^th term = 20
11^thterm = 40
Let the first term of the AP be a and common difference be d
T₇ = a + (n – 1)d = 20
T₇ = a + (7 – 1)d = 20
T₇ = a + 6d = 20 (1)
T₁₁ = a + (n – 1)d = 40
T₁₁ = a + (11 – 1)d = 40
T₁₁ = a + 10d = 40 (2)
Subtracting (1) from (2)
We get,
a + 10d – (a + 6d) = 40 – 20
4d = 20
d = 5

10. What will be the 14^th term of the AP 5, 8, 11, 14, 17…….?
a) 30
b) 41
c) 40
d) 44
View Answer

Answer: d
Explanation: Here a = 5, d = 8 – 5 = 3 and n = 14
T₁₄ = a + (n – 1)d
T₁₄ = 5 + (14 – 1)3
T₁₄ = 5 + 13 × 3
T₁₄ = 44

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