Class 11 Maths Chapter 8 Sequences and Series – Extra Questions with Answers

Sequences and Series problems often come down to spotting whether a list of numbers follows a constant difference or a constant ratio. This set of Class 11 Maths Chapter 8 questions works through geometric progressions, common ratios, and geometric means with worked solutions.

Last Updated: September 23, 2026

Very Short Answer Questions (1 mark)

Q1. Find the common ratio of the GP: 3, 6, 12, 24.
Ans: r = 6/3 = 2.

Q2. Find the 5th term of the GP with a=2, r=3.
Ans: a₅ = 2(3)⁴ = 2(81) = 162.

Q3. Find the sum of the first 5 natural numbers using the formula.
Ans: 5(6)/2 = 15.

Q4. Find the geometric mean of 4 and 16.
Ans: √(4×16) = √64 = 8.

Q5. Does the infinite GP with r=2 have a finite sum?
Ans: No, since |r|=2 is not less than 1.

Short Answer Questions (2–3 marks)

Q6. Find the sum of the infinite GP: 1, 1/2, 1/4, 1/8, …
Ans: a=1, r=1/2 (|r|<1). S∞ = a/(1−r) = 1/(1−0.5) = 1/0.5 = 2.

Q7. Find the arithmetic mean and geometric mean of 9 and 16, and verify that AM ≥ GM.
Ans: AM = (9+16)/2 = 12.5. GM = √(9×16) = √144 = 12. Since 12.5 > 12, AM ≥ GM is confirmed.

Q8. A GP has first term 5 and common ratio 3. Find the sum of the first 4 terms using the sum formula.
Ans: S₄ = a(r⁴−1)/(r−1) = 5(3⁴−1)/(3−1) = 5(81−1)/2 = 5(80)/2 = 200.

Higher-Order Thinking / Application Questions

Q9. A ball is dropped from a height of 10 m, and after each bounce it rises to 3/5 of its previous height. Using the concept of an infinite GP, find the total distance the ball travels (theoretically, until it stops bouncing), accounting for both falling and rising distances.
Ans: The ball falls 10 m initially. Then it bounces up and down repeatedly, each time to 3/5 of the previous height, forming a GP with a=10×(3/5)=6 (first bounce height) and r=3/5, but each bounce height is travelled twice (up and down), except we count the initial fall once. Total distance = 10 (initial fall) + 2 × [sum of infinite GP with a=6, r=3/5] = 10 + 2 × [6/(1−0.6)] = 10 + 2 × 15 = 10 + 30 = 40 m.

Q10. Explain why the formula for the sum of an infinite GP, S∞ = a/(1−r), only works when |r| < 1, using the finite sum formula as a starting point.
Ans: The finite sum formula is S_n = a(1−rⁿ)/(1−r). As n approaches infinity, the term rⁿ approaches 0 only if |r| < 1 (since a fraction raised to higher and higher powers shrinks toward zero). If |r| ≥ 1, rⁿ either stays constant (r=1), grows without bound (r>1), or oscillates without settling (r≤−1), so the sum does not converge to a finite value, and the infinite sum formula does not apply.

Frequently Asked Questions

If the 3rd term of an AP is 8 and the 7th term is 20, how would you find the common difference?
Subtract to get 4d = 12, so d = 3, and this also gives the first term as 2.

How would you find the sum of the first 10 terms of a GP with first term 3 and common ratio 2?
Use the formula Sn = a(r to the n minus 1) over (r minus 1), giving 3(2 to the 10 minus 1) over 1 = 3069.

Written by Satish

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