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Sequences and Series MCQs

Class 11 Maths - questions with answers and worked explanations.

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These questions are drawn from the Pariksha Sutra question bank for Sequences and Series, part of the Class 11 Maths syllabus. Each one shows the correct answer and, where a method helps, the working behind it.

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Questions

Question 1
If the nth term of a sequence is a_n = n^2 + 1, then the 4th term is:
  1. 15
  2. 18
  3. 16
  4. 17
Answer: 17
Explanation

Substitute n = 4 in the given formula a_n = n^2 + 1.

Compute 4^2 = 16 and add 1 to get 16 + 1 = 17.

Thus the 4th term equals 17, which matches the listed answer.

Question 2
The number of terms in the AP 7, 13, 19, ..., 205 is:
  1. 35
  2. 34
  3. 32
  4. 33
Answer: 34
Explanation

Common difference d = 13‑7 = 6.

General term an = a1 + (n‑1)d = 7 + (n‑1)·6.

Set an = 205 → 7 + 6(n‑1) = 205 → 6(n‑1) = 198 → n‑1 = 33 → n = 34.

Hence the AP has 34 terms.

Question 3
The sum of the first n even numbers 2 + 4 + 6 + ... equals:
  1. n(n+1)
  2. n^2
  3. 2n
  4. n(n+1)/2
Answer: n(n+1)
Explanation

Even numbers form an AP with a=2, d=2; sum of first n terms = n/2[2a+(n-1)d] = n/2[4+2(n-1)] = n/2[2n+2] = n(n+1). Hence the sum equals n(n+1).

Question 4
The arithmetic mean of the first n natural numbers is:
  1. (n+1)/2
  2. n(n+1)/2
  3. n/2
  4. (n-1)/2
Answer: (n+1)/2
Explanation

Sum of first n natural numbers = n(n+1)/2.

Arithmetic mean = (sum)/n = [n(n+1)/2] ÷ n = (n+1)/2.

Thus the mean of the first n natural numbers is (n+1)/2.

Question 5
The sum of the GP 1 + 2 + 4 + ... up to 10 terms is:
  1. 1023
  2. 2047
  3. 1000
  4. 511
Answer: 1023
Explanation

First term a=1, common ratio r=2, number of terms n=10.

Sum of a GP = a(r^n−1)/(r−1) = 1(2^{10}−1)/(2−1).

2^{10}=1024, so sum = (1024−1)=1023.

Hence the correct answer is 1023.

Question 6
The sum of the first n odd numbers 1 + 3 + 5 + ... + (2n-1) equals:
  1. n(n+1)/2
  2. n^2
  3. n(n+1)
  4. 2n
Answer: n^2
Explanation

Add the terms as an arithmetic series: first term a=1, common difference d=2, number of terms n.

Sum formula S=n/2[2a+(n-1)d]=n/2[2·1+(n-1)·2]=n/2[2+2n-2]=n/2·2n=n².

Thus the sum of the first n odd numbers equals n².

Question 7
If a, b, c are in GP, then which relation holds?
  1. b^2 = a + c
  2. b^2 = ac
  3. 2b = a + c
  4. b = a + c
Answer: b^2 = ac
Explanation

For a geometric progression the common ratio r satisfies b = ar and c = ar².

Hence b² = (ar)² = a·ar² = a·c.

Thus the relation that always holds is b² = ac.

Question 8
If the nth term of an AP is 3n - 1, its common difference is:
  1. 2
  2. -1
  3. 3
  4. 1
Answer: 3
Explanation

For an AP, a_n = a_1 + (n‑1)d.

Given a_n = 3n – 1, write a_n = (3·1 – 1) + (n‑1)·d = 2 + (n‑1)d.

Equate coefficients of n: 3 = d, and constant terms give 2 = 2, confirming d = 3.

Hence the common difference is 3.

Question 9
For the AP 3, 7, 11, ..., the 20th term is:
  1. 77
  2. 75
  3. 83
  4. 79
Answer: 79
Explanation

Common difference d = 7−3 = 4.

General term of an AP: aₙ = a₁ + (n−1)d.

For n = 20: a₂₀ = 3 + (20−1)·4 = 3 + 76 = 79.

Hence the 20th term is 79.

Question 10
If 5, x, 17 are in AP, then x equals:
  1. 10
  2. 12
  3. 9
  4. 11
Answer: 11
Explanation

In an arithmetic progression, the middle term equals the average of the two extremes.

So \(x = \frac{5+17}{2}\).

\(\frac{5+17}{2}= \frac{22}{2}=11\).

Hence x equals 11.

Question 11
In an AP with common difference 4, the value of a_10 - a_6 is:
  1. 20
  2. 16
  3. 12
  4. 8
Answer: 16
Explanation

For an AP, a_n = a_1 + (n‑1)d.

Thus a_10 – a_6 = [a_1 + 9d] – [a_1 + 5d] = (9d – 5d) = 4d.

Given d = 4, substitute: 4 × 4 = 16.

Hence the answer is 16.

Question 12
The value of 51 + 52 + 53 + ... + 100 is:
  1. 3675
  2. 3750
  3. 3800
  4. 3775
Answer: 3775
Explanation

Number of terms from 51 to 100 inclusive = 100‑51+1 = 50.

Sum of an arithmetic series = (number of terms) × (first + last) ÷ 2.

So sum = 50 × (51 + 100) ÷ 2 = 50 × 151 ÷ 2 = 25 × 151 = 3775.

Hence the correct answer is 3775.

Question 13
The arithmetic mean of 8 and 20 is:
  1. 14
  2. 16
  3. 12
  4. 13
Answer: 14
Explanation

Arithmetic mean = (first term + second term)/2

= (8 + 20)/2 = 28/2 = 14

Thus the arithmetic mean of 8 and 20 is 14.

Question 14
The geometric mean inserted between 3 and 27 is:
  1. 18
  2. 9
  3. 6
  4. 12
Answer: 9
Explanation

Let the geometric mean be \(g\). For three numbers in G.P., \(3, g, 27\) we have \(g^2 = 3 \times 27\).

Thus \(g^2 = 81\) and \(g = \sqrt{81}=9\).

Hence the inserted geometric mean is 9.

Question 15
In a GP the 7th term is 8 times the 4th term. The common ratio is:
  1. 4
  2. 8
  3. 3
  4. 2
Answer: 2
Explanation

For a GP, \(a_n = a_1 r^{\,n-1}\).

Given \(a_7 = 8\,a_4\) → \(a_1 r^{6}=8\,a_1 r^{3}\).

Cancel \(a_1\) and divide by \(r^{3}\): \(r^{3}=8\).

Thus \(r = \sqrt[3]{8}=2, which matches the given answer.

Question 16
The common ratio of an infinite GP with first term 5 and sum 15 is:
  1. 1/2
  2. 3/4
  3. 2/3
  4. 1/3
Answer: 2/3
Explanation

Sum of infinite GP = a/(1‑r) → 15 = 5/(1‑r)

Cross‑multiply: 15(1‑r)=5 → 15‑15r=5 → 15r=10 → r=10/15 = 2/3

Thus the common ratio is 2/3.

Question 17
For the GP 2, 6, 18, ..., the 5th term is:
  1. 162
  2. 108
  3. 54
  4. 486
Answer: 162
Explanation

Common ratio r = 6/2 = 3.

General term of GP: aₙ = a₁·r^(n‑1).

For n = 5: a₅ = 2·3^(5‑1) = 2·81 = 162.

Hence the 5th term is 162.

Question 18
The sum of the infinite GP 1 + 1/2 + 1/4 + ... is:
  1. 2
  2. 3/2
  3. 1
  4. 4
Answer: 2
Explanation

First term a = 1 and common ratio r = ½ (|r|<1, so series converges).

Sum of an infinite GP = a/(1−r).

Substituting gives S = 1/(1−½) = 1/(½) = 2.

Hence the sum is 2.

Question 19
The sum of the series 2 + 4 + 6 + ... + 100 is:
  1. 2550
  2. 2600
  3. 2450
  4. 2500
Answer: 2550
Explanation

Number of terms n = 100/2 = 50.

Sum of AP = n/2 × (first + last) = 50/2 × (2 + 100).

Compute: 25 × 102 = 2550.

Hence the sum is 2550.

Question 20
If three arithmetic means are inserted between 1 and 9, the second mean is:
  1. 3
  2. 6
  3. 4
  4. 5
Answer: 5
Explanation

Let the common difference be d. With three means there are 5 terms: 1, a₂, a₃, a₄, 9. So 9 = 1 + 4d ⇒ d = 2.

The terms are 1, 3, 5, 7, 9; the second inserted mean is a₃ = 5.

Hence the answer is 5.

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