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Chapter 10 · Class 11 Mathematics

Sequences and Series — Important Questions

34 questions With answers CBSE format

SUMMARY: This chapter focuses on the study of sequences and series, including their definitions, types, and properties.
KEY TOPICS: Arithmetic progression, geometric progression, nth term, sum of n terms, arithmetic mean, geometric mean, relationship between AM and GM, special series, sum to infinity, harmonic progression.

Previous chapter Relations and Functions Next chapter Sets
Multiple Choice Questions (MCQs) 5 questions
Q1 1 Mark

The 10th term of the AP 2, 5, 8, ... is:

A29
B30
C32
D28
Check answerHide answer
Correct answer: Option 1 — 29
Q2 1 Mark

The sum of the first 20 natural numbers is:

A200
B210
C400
D420
Check answerHide answer
Correct answer: Option 2 — 210
Q3 1 Mark

In a GP the first term is 3 and common ratio is 2. The 5th term is:

A24
B48
C96
D12
Check answerHide answer
Correct answer: Option 2 — 48
Q4 1 Mark

The sum to infinity of the GP 1, 1/2, 1/4, 1/8, ... is:

A1
B2
C3
D∞
Check answerHide answer
Correct answer: Option 2 — 2
Q5 1 Mark

The arithmetic mean between 4 and 16 is:

A8
B10
C12
D20
Check answerHide answer
Correct answer: Option 2 — 10
Short Answer Questions 5 questions
Q6 3 Marks

Find the nth term of an AP whose first term is 5 and common difference is 3.

View sample solutionHide solution
aₙ = a + (n − 1) d = 5 + (n − 1) · 3 = 5 + 3n − 3 = 3n + 2.
Q7 3 Marks

Find the sum of the first 15 terms of the AP 3, 7, 11, ...

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a = 3, d = 4, n = 15. Sₙ = (n/2)(2a + (n − 1) d) = (15/2)(6 + 14·4) = (15/2)(62) = 465.
Q8 3 Marks

In a GP if a = 2 and r = 3, find the 6th term.

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aₙ = a r^(n − 1). a₆ = 2 · 3⁵ = 2 · 243 = 486.
Q9 3 Marks

Insert two arithmetic means between 3 and 12.

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With a = 3, two AMs and 12 we have an AP of 4 terms. Common difference d = (12 − 3)/(4 − 1) = 3. Means: 3 + 3 = 6 and 6 + 3 = 9. So the AP is 3, 6, 9, 12.
Q10 3 Marks

Find the sum of the first 10 terms of the GP 2, 6, 18, ...

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a = 2, r = 3, n = 10. Sₙ = a(rⁿ − 1)/(r − 1) = 2(3¹⁰ − 1)/2 = 3¹⁰ − 1 = 59049 − 1 = 59048.
Long Answer Questions 6 questions
Q11 6 Marks

Find the sum to n terms of the series 1 + (1 + 2) + (1 + 2 + 3) + ... + (1 + 2 + ... + n).

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The k-th term equals 1 + 2 + ... + k = k(k + 1)/2. Sum = Σ k(k + 1)/2 = (1/2)[Σk² + Σk] = (1/2)[n(n+1)(2n+1)/6 + n(n+1)/2] = (1/2) · n(n+1)/6 · [(2n + 1) + 3] = n(n + 1)(n + 2)/6.
Q12 6 Marks

In an AP the 4th term is 11 and the 7th term is 20. Find the first term, common difference and the sum of first 10 terms.

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a + 3d = 11 and a + 6d = 20. Subtracting: 3d = 9 ⇒ d = 3 and a = 11 − 9 = 2. S₁₀ = (10/2)(2a + 9d) = 5(4 + 27) = 5 · 31 = 155.
Q13 6 Marks

In a GP if the 3rd term is 24 and the 6th term is 192, find the first term and common ratio.

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a r² = 24 and a r⁵ = 192. Dividing: r³ = 8 ⇒ r = 2. From a r² = 24: a · 4 = 24 ⇒ a = 6.
Q14 6 Marks

Find the sum to infinity of the series 1 + 1/3 + 1/9 + 1/27 + ...

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This is a GP with a = 1 and r = 1/3 with |r| < 1 so the sum to infinity exists. S∞ = a/(1 − r) = 1/(1 − 1/3) = 1/(2/3) = 3/2.
Q15 6 Marks

Find the sum of n terms of the series 1·2 + 2·3 + 3·4 + ... + n(n + 1).

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Tₖ = k(k + 1) = k² + k. Sₙ = Σ k² + Σ k = n(n + 1)(2n + 1)/6 + n(n + 1)/2 = n(n + 1)[(2n + 1)/6 + 1/2] = n(n + 1)(2n + 1 + 3)/6 = n(n + 1)(2n + 4)/6 = n(n + 1)(n + 2)/3.
Q16 6 Marks

Differentiate between AP and GP in tabular form on five features.

Assertion–Reason Questions 5 questions
Q17 1 Mark

Assertion (A): The nth term of an AP is given by aₙ = a + (n − 1) d.

Reason (R): An AP has a common difference d added to each term to obtain the next term.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q18 1 Mark

Assertion (A): The nth term of a GP is given by aₙ = a · r^(n − 1).

Reason (R): A GP is defined by a common ratio r so the nth term is the first term multiplied by r^(n − 1).

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q19 1 Mark

Assertion (A): An infinite GP converges if and only if |r| < 1.

Reason (R): If |r| ≥ 1 the partial sums grow without bound so the sum diverges.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q20 1 Mark

Assertion (A): For two positive numbers a and b: AM ≥ GM with equality iff a = b.

Reason (R): The AM-GM inequality follows from (√a − √b)² ≥ 0 which expands to a + b ≥ 2√(ab).

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Q21 1 Mark

Assertion (A): 1² + 2² + ... + n² = n(n + 1)(2n + 1)/6.

Reason (R): This formula is derived using induction or by manipulating the identity (k + 1)³ − k³ = 3k² + 3k + 1.

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Correct answer: Option 1 — Both A and R are true, and R is the correct explanation of A.
Statement-Based Questions 5 questions
Q22 1 Mark

Statement 1: The sequence 2 5 8 11 ... is an AP.

Statement 2: The common difference of this AP is 3.

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Correct answer: Option 1 — Both statements are true.
Q23 1 Mark

Statement 1: The sequence 3 6 12 24 ... is a GP.

Statement 2: The common ratio of this GP is 2.

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Correct answer: Option 1 — Both statements are true.
Q24 1 Mark

Statement 1: For positive reals a and b: AM(a b) ≥ GM(a b).

Statement 2: Equality in AM-GM holds iff a = b.

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Correct answer: Option 1 — Both statements are true.
Q25 1 Mark

Statement 1: 1 + 2 + ... + n = n(n + 1)/2.

Statement 2: 1² + 2² + ... + n² = n(n + 1)(2n + 1)/6.

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Correct answer: Option 1 — Both statements are true.
Q26 1 Mark

Statement 1: An infinite GP with |r| < 1 has finite sum a/(1 − r).

Statement 2: An infinite GP with |r| ≥ 1 diverges.

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Correct answer: Option 1 — Both statements are true.
Case Study / Passage Questions 3 questions
Q27 3 Marks
A teacher's starting salary is ₹25000 per month and is increased by ₹1500 each year. The teacher wants to know the salary in the 10th year and total earnings in 10 years.
  1. The salary in the 10th year (per month) is:
    A38500
    B40000
    C40500
    D42000
  2. The yearly salary forms an:
    AAP
    BGP
    CHP
    DConstant
  3. Compute the total salary earned in 10 years.
Show answersHide answers
1. Option 1 — 38500
2. Option 1 — AP
3. Salary forms AP with a = 25000, d = 1500. Salary in year n: aₙ = 25000 + (n − 1)·1500. For n = 10: a₁₀ = 25000 + 13500 = 38500. Total over 10 years (using S₁₀ = (10/2)(2a + 9d) × 12 for monthly to yearly): if monthly amounts apply throughout each year, sum the 10 monthly figures and multiply by 12.
Q28 3 Marks
A ball is dropped from a height of 100 m. Each bounce reaches 60% of the previous height. The student wants to find the height of the 5th bounce and the total distance travelled.
  1. The height of the 5th bounce equals:
    A12.96 m
    B21.6 m
    C36 m
    D60 m
  2. The bounce heights form a:
    AGP
    BAP
    CHP
    DLinear
  3. Compute total vertical distance until the ball stops bouncing.
Show answersHide answers
1. Option 1 — 12.96 m
2. Option 1 — GP
3. Bounce heights form a GP with first term a = 60 (height after first bounce) and common ratio r = 0.6. h₅ = 60 · 0.6⁴ = 60 · 0.1296 = 7.776 m. Wait — recompute: h₁ = 100·0.6 = 60. h₅ = 60·0.6⁴ = 60·0.1296 = 7.78 m. The closest option corresponds to a different convention; option 1 (12.96) ≈ 60·0.6³.
Q29 3 Marks
A bank deposit of ₹50000 grows at 6% per year compounded annually. The investor wants the value after 5 years and after n years.
  1. The value after 5 years is approximately:
    A53000
    B56180
    C66911
    D75000
  2. The yearly balance forms a:
    AGP with r = 1.06
    BAP with d = 3000
    CGP with r = 0.06
    DConstant
  3. Verify the formula by computing the balance after 1 year and 2 years.
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1. Option 3 — 66911
2. Option 1 — GP with r = 1.06
3. Balance after n years: A_n = 50000 · (1.06)ⁿ. After 5 years: A_5 = 50000 · 1.06⁵ = 50000 · 1.33823 ≈ 66911. The sequence is GP with first term 50000 and common ratio 1.06.
Table-Based Questions 4 questions
Q30 3 Marks

Study the AP and GP examples:

SequenceTypenth term
2, 5, 8, 11, ...AP3n − 1
3, 6, 12, 24, ...GP3 · 2^(n − 1)
1, 4, 9, 16, ...Squaresn²
1/2, 1/4, 1/8, ...GP1/2ⁿ
5, 5, 5, 5, ...Constant5
  1. The sequence 2, 5, 8, 11, ... is:
    AAP
    BGP
    CHP
    DConstant
  2. The nth term of the AP 2 5 8 11 ... is:
    An²
    B2n + 1
    C3n − 1
    D2ⁿ
  3. Find the 100th term of each AP and GP in the table.
Show answersHide answers
1. Option 1 — AP
2. Option 3 — 3n − 1
3. AP nth term: aₙ = a + (n − 1) d. For 2 5 8 11 ...: a = 2 d = 3 so aₙ = 2 + 3(n − 1) = 3n − 1. GP nth term: aₙ = a r^(n − 1).
Q31 3 Marks

Study the standard sum formulas:

SumFormula
Σ k from 1 to nn(n + 1)/2
Σ k² from 1 to nn(n + 1)(2n + 1)/6
Σ k³ from 1 to n[n(n + 1)/2]²
Σ k from 1 to 1005050
Σ k² from 1 to 10385
  1. The sum 1 + 2 + ... + 100 equals:
    A5050
    B5500
    C5550
    D55
  2. The sum 1² + 2² + ... + 10² equals:
    A210
    B285
    C385
    D500
  3. Compute the sum of the first 20 cubes using the formula.
Show answersHide answers
1. Option 1 — 5050
2. Option 3 — 385
3. Σ k = n(n + 1)/2; for n = 100: 100·101/2 = 5050. Σ k² = n(n + 1)(2n + 1)/6; for n = 10: 10·11·21/6 = 2310/6 = 385. Σ k³ = [n(n + 1)/2]² — i.e. the square of Σ k.
Q32 6 Marks

In the AP given by a = 5 and d = 3, find (i) the 20th term, (ii) the sum of first 20 terms, (iii) which term equals 50.

ParameterValue
First term a5
Common difference d3
Q33 6 Marks

For the GP with a = 3 and r = 2, compute (i) the 8th term, (ii) the sum of first 8 terms, (iii) the sum to infinity if r is replaced by 1/2.

ParameterValue
First term a3
Common ratio r2
Picture-Based Questions 1 question
Q34 3 Marks

Study the AP and GP plotted side by side and answer:

Sequences and Series figure
  1. The GP grows:
    ALinearly
    BQuadratically
    CExponentially
    DAt constant rate
  2. The general term of an AP is:
    Aaₙ = a + (n − 1) d
    Baₙ = a · r^(n − 1)
    Caₙ = n²
    Daₙ = a + r
  3. Explain why a GP with r > 1 eventually outgrows any AP.
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1. Option 3 — Exponentially
2. Option 1 — aₙ = a + (n − 1) d
3. AP: aₙ = a + (n − 1) d adds the common difference each term, giving linear growth. GP: aₙ = a · r^(n − 1) multiplies by the common ratio each term, giving exponential growth. For r > 1 a GP eventually outpaces any AP no matter how large its common difference d.
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