Sequences and Series — Class 11 MCQs with Answers

Sequences and Series — important questions & MCQs with answers (Class 11 Mathematics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   nth term of AP a, a+d, a+2d, ... :

   AndBa + ndCa·dⁿ⁻¹Da + (n−1)d
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   Answer: a + (n−1)d

   Explanation: The nth term of an AP is a + (n−1)d, not a + nd — count n−1 steps from the first term, since a is already term 1 before any step of size d is added.

2. Q2 of 6Easy

   In GP 2, 6, 18, ..., common ratio:

   A4B3C2D6
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   Answer: 3

   Explanation: The common ratio is each term divided by the one before: 6/2 = 3 (also 18/6 = 3). Confusing it with the difference 6−2 = 4 gives the wrong AP-style answer.

3. Q3 of 6Medium

   Three arithmetic means are inserted between 3 and 19. The second mean is:

   A15B9C11D7
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   Answer: 11

   Explanation: The five terms have common difference (19 − 3)/4 = 4, giving 3, 7, 11, 15, 19.

4. Q4 of 6Medium

   Assertion (A): For the positive numbers 4 and 9, the arithmetic mean is less than the geometric mean. Reason (R): For any two positive real numbers, A.M. ≥ G.M.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: A is false but R is true

   Explanation: A is false: A.M. = (4 + 9)/2 = 6.5 and G.M. = √36 = 6, so the A.M. is greater. R is true; it is the A.M.–G.M. relation A ≥ G.

5. Q5 of 6Hard

   Σ(k=1 to 100) (k² + k) =

   A343400B338350C5050D505000
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   Answer: 343400

   Explanation: Σk² = 100·101·201/6 = 338350 and Σk = 100·101/2 = 5050, so Σ(k²+k) = 343400. Adding only Σk² or only Σk instead of both misses half the total.

6. Q6 of 6Hard

   Assertion (A): If the A.M. and G.M. of the roots of a quadratic equation are 8 and 5 respectively, then the equation is x² – 8x + 25 = 0. Reason (R): A quadratic equation whose roots have sum S and product P is x² – Sx + P = 0.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: A is false but R is true

   Explanation: A is false: A.M. = 8 gives sum of roots = 16, and G.M. = 5 gives product = 25, so the equation is x² – 16x + 25 = 0. R is true; using it with S = 16 and P = 25 is what shows A is wrong.

Sequences and Series notes: key concepts (Class 11 Mathematics)

A sequence is an ordered list of numbers and a series is their sum. This chapter studies arithmetic progressions with a common difference, geometric progressions with a common ratio, their nth terms and sums, the arithmetic and geometric means with the AM ≥ GM inequality, and the special sums Σn, Σn² and Σn³.

Sequence and series

A sequence lists terms in order by a rule; a series adds those terms. Finite sequences end, while infinite ones continue without a last term.

Arithmetic progression (AP)

Each term exceeds the previous by a fixed common difference d; the nth term is aₙ = a + (n−1)d, which is linear in n.

Sum of an AP

Sₙ = n/2[2a + (n−1)d] = n/2(a + l), where l is the last term; the sum grows quadratically with the number of terms n.

Arithmetic mean

The AM of a and b is (a + b)/2; inserting k arithmetic means between two numbers builds an AP with k + 2 terms in total.

Geometric progression (GP)

Each term is the previous multiplied by a fixed common ratio r; the nth term is aₙ = a r^(n−1), growing or decaying geometrically.

Sum of a GP

Sₙ = a(rⁿ − 1)/(r − 1) holds when r ≠ 1; if the ratio r = 1 every term is equal and the sum collapses to simply na.

Infinite GP

If |r| < 1 the terms shrink and the sum converges to S∞ = a/(1 − r); if |r| ≥ 1 the series has no finite sum.

Geometric mean

The GM of positive a and b is √(ab); inserting k geometric means between two numbers builds a GP with k + 2 terms in total.

AM–GM inequality

For positive numbers the arithmetic mean is at least the geometric mean, AM ≥ GM, with equality only when all the numbers are equal.

Sum of first n naturals

Σn = n(n + 1)/2 adds 1 + 2 + … + n; it is itself an AP sum and recurs throughout series and counting problems.

Sums of squares and cubes

Σn² = n(n+1)(2n+1)/6 and Σn³ = [n(n+1)/2]², so the sum of cubes equals the square of the sum of the first n naturals.

Special series

Split a general term into standard pieces (Σn, Σn², constants), or use partial fractions to telescope and sum non-standard series.