ClassApna

Class 11 Maths NCERT Solutions

~11 min read

Sequences and Series Class 11 Maths NCERT Solutions

The complete NCERT exercise solutions for Chapter 8, Sequences and Series — 46 questions from Ex 8.1 to Ex 8.2, each worked through step by step in the CBSE marking pattern. Arithmetic and geometric progressions, the sum formulas, and the standard sums of n, n² and n³.

Class:11Subject:MathsChapter:8
4 Key Formulas
DWritten byDeep Narayan
Updated
Key Concept Summary

How many questions are in NCERT Class 11 Maths Chapter 8?

Chapter 8 carries 2 exercise questions, numbered Ex 8.1 to Ex 8.2. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

01

Chapter Overview

A sequence lists numbers in a fixed order; its series is the running sum. Two progressions dominate the chapter: the arithmetic progression (AP), where each term grows by a constant difference d, and the geometric progression (GP), where each term multiplies by a constant ratio r. Their nth terms and sums, plus the arithmetic and geometric means (A.M. and G.M.), power the word problems at the end.

Board pattern

Quote the formula in symbols before substituting: aₙ = a + (n−1)d, Sₙ = n/2 (2a + (n−1)d), aₙ = arⁿ⁻¹, Sₙ = a(rⁿ−1)/(r−1). For instalment and depreciation questions, identify the AP or GP hidden inside the story and state the first term and common difference/ratio explicitly. For a GP that forms an AP after modification, write both progressions before solving.
02

Exercise 8.1 — Sequences

14Exercise questions

Step-by-step solution

  1. 1a₁ = 1(3) = 3, a₂ = 2(4) = 8, a₃ = 3(5) = 15.
  2. 2a₄ = 4(6) = 24, a₅ = 5(7) = 35.

Final answer

3, 8, 15, 24, 35.

Step-by-step solution

  1. 1a₁ = 1/2, a₂ = 2/3, a₃ = 3/4.
  2. 2a₄ = 4/5, a₅ = 5/6.

Final answer

1/2, 2/3, 3/4, 4/5, 5/6.

Step-by-step solution

  1. 1a₁ = 2, a₂ = 4, a₃ = 8.
  2. 2a₄ = 16, a₅ = 32.

Final answer

2, 4, 8, 16, 32.

Step-by-step solution

  1. 1a₁ = −1/6, a₂ = 1/6, a₃ = 3/6 = 1/2.
  2. 2a₄ = 5/6, a₅ = 7/6.

Final answer

−1/6, 1/6, 1/2, 5/6, 7/6.

Step-by-step solution

  1. 1a₁ = (−1)⁰·5² = 25, a₂ = (−1)¹·5³ = −125.
  2. 2a₃ = 625, a₄ = −3125, a₅ = 15625.

Final answer

25, −125, 625, −3125, 15625.

Step-by-step solution

  1. 1a₁ = 1(6)/4 = 3/2, a₂ = 2(9)/4 = 9/2, a₃ = 3(14)/4 = 21/2.
  2. 2a₄ = 4(21)/4 = 21, a₅ = 5(30)/4 = 75/2.

Final answer

3/2, 9/2, 21/2, 21, 75/2.

Step-by-step solution

  1. 1a₁₇ = 4(17) − 3 = 68 − 3 = 65.
  2. 2a₂₄ = 4(24) − 3 = 96 − 3 = 93.

Final answer

65, 93.

Step-by-step solution

  1. 1a₇ = 7²/2⁷ = 49/128.

Final answer

49/128.

Step-by-step solution

  1. 1a₉ = (−1)⁸·9³ = 729.

Final answer

729.

Step-by-step solution

  1. 1a₂₀ = 20(20 − 2)/(20 + 3) = 20 × 18/23 = 360/23.

Final answer

360/23.

Step-by-step solution

  1. 1a₁ = 3, a₂ = 3(3) + 2 = 11, a₃ = 3(11) + 2 = 35.
  2. 2a₄ = 3(35) + 2 = 107, a₅ = 3(107) + 2 = 323.
  3. 3Series: 3 + 11 + 35 + 107 + 323 + …

Final answer

3, 11, 35, 107, 323; series 3 + 11 + 35 + 107 + 323 + ….

Step-by-step solution

  1. 1a₁ = −1, a₂ = −1/2, a₃ = (−1/2)/3 = −1/6.
  2. 2a₄ = (−1/6)/4 = −1/24, a₅ = (−1/24)/5 = −1/120.
  3. 3Series: −1 − 1/2 − 1/6 − 1/24 − 1/120.

Final answer

−1, −1/2, −1/6, −1/24, −1/120; series −1 − 1/2 − 1/6 − 1/24 − 1/120.

Step-by-step solution

  1. 1a₃ = a₂ − 1 = 1, a₄ = a₃ − 1 = 0, a₅ = a₄ − 1 = −1.
  2. 2Series: 2 + 2 + 1 + 0 + (−1).

Final answer

2, 2, 1, 0, −1; series 2 + 2 + 1 + 0 − 1.

Step-by-step solution

  1. 1a₃ = 1 + 1 = 2, a₄ = 2 + 1 = 3, a₅ = 3 + 2 = 5, a₆ = 5 + 3 = 8.
  2. 2n = 1: a₂/a₁ = 1; n = 2: a₃/a₂ = 2.
  3. 3n = 3: a₄/a₃ = 3/2; n = 4: a₅/a₄ = 5/3; n = 5: a₆/a₅ = 8/5.

Final answer

1, 2, 3/2, 5/3, 8/5.

03

Exercise 8.2 — Geometric Progressions

32Exercise questions

Step-by-step solution

  1. 1a = 5/2, r = (5/4)/(5/2) = 1/2.
  2. 2a₂₀ = (5/2)(1/2)¹⁹ = 5/2²⁰.
  3. 3aₙ = (5/2)(1/2)ⁿ⁻¹ = 5/2ⁿ.

Final answer

5/2²⁰ and 5/2ⁿ.

Step-by-step solution

  1. 1a₈ = ar⁷ = a·2⁷ = 128a = 192 → a = 3/2.
  2. 2a₁₂ = (3/2)·2¹¹ = (3/2)(2048) = 3072.

Final answer

3072.

Step-by-step solution

  1. 1p = ar⁴, q = ar⁷, s = ar¹⁰.
  2. 2q² = a²r¹⁴ and ps = (ar⁴)(ar¹⁰) = a²r¹⁴.
  3. 3Hence q² = ps.

Final answer

q² = ps (shown).

Step-by-step solution

  1. 1Let terms be a, ar, ar², ar³ with a = −3.
  2. 24th = (2nd)² ⇒ ar³ = (ar)² ⇒ r = a = −3.
  3. 37th term = ar⁶ = (−3)(−3)⁶ = (−3)⁷ = −2187.

Final answer

−2187.

Step-by-step solution

  1. 1(i) a = 2, r = √2; 2·(√2)ⁿ⁻¹ = 128 = 2⁷ → (√2)ⁿ⁻¹ = 2⁶ → (n−1)/2 = 6 → n = 13.
  2. 2(ii) a = √3, r = √3; (√3)ⁿ = 729 = 3⁶ → 3ⁿᐟ² = 3⁶ → n = 12.
  3. 3(iii) a = r = 1/3; (1/3)ⁿ = 1/19683 = (1/3)⁹ → n = 9.

Final answer

(i) 13th, (ii) 12th, (iii) 9th.

Step-by-step solution

  1. 1x² = (−2/7)(−7/2) = 1.
  2. 2x = ±1.

Final answer

x = ±1.

Step-by-step solution

  1. 1a = 0.15, r = 0.1.
  2. 2S₂₀ = 0.15(1 − (0.1)²⁰)/(1 − 0.1) = (1/6)(1 − 10⁻²⁰).

Final answer

(1/6)(1 − 10⁻²⁰).

Step-by-step solution

  1. 1a = √7, r = √21/√7 = √3.
  2. 2Sₙ = √7[(√3)ⁿ − 1]/(√3 − 1).

Final answer

√7(3ⁿᐟ² − 1)/(√3 − 1).

Step-by-step solution

  1. 1a₁ = 1, r = −a.
  2. 2Sₙ = 1(1 − (−a)ⁿ)/(1 − (−a)) = (1 − (−a)ⁿ)/(1 + a).

Final answer

(1 − (−a)ⁿ)/(1 + a).

Step-by-step solution

  1. 1a = x³, r = x².
  2. 2Sₙ = x³(x²ⁿ − 1)/(x² − 1).

Final answer

x³(x²ⁿ − 1)/(x² − 1).

Step-by-step solution

  1. 1Σ(2 + 3ᵏ) = 2·11 + (3 + 3² + … + 3¹¹).
  2. 2= 22 + 3(3¹¹ − 1)/(3 − 1) = 22 + (3¹² − 3)/2.
  3. 3= 22 + (531441 − 3)/2 = 22 + 265719 = 265741.

Final answer

265741.

Step-by-step solution

  1. 1Let the terms be a/r, a, ar; product a³ = 1 → a = 1.
  2. 2Sum: 1/r + 1 + r = 39/10 → 1/r + r = 29/10 → 10r² − 29r + 10 = 0.
  3. 3r = (29 ± 21)/20 → r = 5/2 or r = 2/5.
  4. 4Terms: (2/5, 1, 5/2) or (5/2, 1, 2/5).

Final answer

r = 5/2 or 2/5; terms (2/5, 1, 5/2) or (5/2, 1, 2/5).

Step-by-step solution

  1. 1a = 3, r = 3, Sₙ = 3(3ⁿ − 1)/(3 − 1) = 3(3ⁿ − 1)/2.
  2. 23(3ⁿ − 1)/2 = 120 → 3ⁿ − 1 = 80 → 3ⁿ = 81 = 3⁴.
  3. 3n = 4.

Final answer

4 terms.

Step-by-step solution

  1. 1First three sum: a(1 + r + r²) = 16.
  2. 2Next three: ar³(1 + r + r²) = 128 ⇒ 16r³ = 128 → r³ = 8 → r = 2.
  3. 3a(1 + 2 + 4) = 16 → 7a = 16 → a = 16/7.
  4. 4Sₙ = (16/7)(2ⁿ − 1)/(2 − 1) = (16/7)(2ⁿ − 1).

Final answer

a = 16/7, r = 2, Sₙ = (16/7)(2ⁿ − 1).

Step-by-step solution

  1. 1ar⁶ = 64 → 729r⁶ = 64 → r⁶ = 64/729 = (2/3)⁶ → r = 2/3.
  2. 2S₇ = 729(1 − (2/3)⁷)/(1 − 2/3) = 729·3(1 − 128/2187).
  3. 3= 2187 − 128 = 2059.

Final answer

2059.

Step-by-step solution

  1. 1a + ar = −4 and ar⁴ = 4ar² → r² = 4 → r = ±2.
  2. 2r = 2: a(1 + 2) = −4 → a = −4/3.
  3. 3r = −2: a(1 − 2) = −4 → a = 4.
  4. 4G.P.s: (−4/3, −8/3, …) or (4, −8, …).

Final answer

a = −4/3 with r = 2, or a = 4 with r = −2.

Step-by-step solution

  1. 1x = ar³, y = ar⁹, z = ar¹⁵.
  2. 2y² = a²r¹⁸ and xz = a²r¹⁸.
  3. 3y² = xz, so x, y, z are in G.P.

Final answer

Proved (y² = xz).

Step-by-step solution

  1. 1kth term = 8(10ᵏ − 1)/9.
  2. 2Sₙ = (8/9)Σ(10ᵏ − 1) = (8/9)[10(10ⁿ − 1)/9 − n].
  3. 3= 80(10ⁿ − 1)/81 − 8n/9.

Final answer

80(10ⁿ − 1)/81 − 8n/9.

Step-by-step solution

  1. 1Products: 2·128 = 256, 4·32 = 128, 8·8 = 64, 16·2 = 32, 32·(1/2) = 16.
  2. 2Sum = 256 + 128 + 64 + 32 + 16 = 496.

Final answer

496.

Step-by-step solution

  1. 1Products are aA, aArR, aAr²R², …, aA(rR)ⁿ⁻¹.
  2. 2Each term is the previous × (rR).
  3. 3They form a G.P. with first term aA and common ratio rR.

Final answer

G.P. with common ratio rR (shown).

Step-by-step solution

  1. 1Let the numbers be a, ar, ar², ar³: ar² − a = 9 and ar − ar³ = 18.
  2. 2a(r² − 1) = 9 and ar(1 − r²) = 18 → −r·a(r² − 1) = 18 → −r·9 = 18 → r = −2.
  3. 3a(4 − 1) = 9 → a = 3.
  4. 4Numbers: 3, −6, 12, −24.

Final answer

3, −6, 12, −24.

Step-by-step solution

  1. 1a = A r^(p−1), b = A r^(q−1), c = A r^(r−1).
  2. 2Exponent of A: (q−r) + (r−p) + (p−q) = 0.
  3. 3Exponent of r: (p−1)(q−r) + (q−1)(r−p) + (r−1)(p−q) = 0.
  4. 4Result: A⁰r⁰ = 1.

Final answer

Proved: the product equals 1.

Step-by-step solution

  1. 1b = arⁿ⁻¹ and P = aⁿr^(n(n−1)/2).
  2. 2P² = a²ⁿr^(n(n−1)) and (ab)ⁿ = (a·arⁿ⁻¹)ⁿ = a²ⁿr^(n(n−1)).
  3. 3P² = (ab)ⁿ.

Final answer

P² = (ab)ⁿ (shown).

Step-by-step solution

  1. 1Sum of first n terms: S = a(1 − rⁿ)/(1 − r).
  2. 2Sum of (n+1)th to (2n)th: arⁿ(1 − rⁿ)/(1 − r).
  3. 3Ratio = [a(1−rⁿ)/(1−r)] / [arⁿ(1−rⁿ)/(1−r)] = 1/rⁿ.

Final answer

Ratio = 1/rⁿ (shown).

Step-by-step solution

  1. 1Let b = ar, c = ar², d = ar³.
  2. 2LHS = a²(1 + r² + r⁴)·a²r²(1 + r² + r⁴) = a⁴r²(1 + r² + r⁴)².
  3. 3RHS = (a²r + a²r³ + a²r⁵)² = a⁴r²(1 + r² + r⁴)².
  4. 4Both sides are equal.

Final answer

Identity proved.

Step-by-step solution

  1. 1Let 3, x, y, 81 be a G.P.: 3r³ = 81 → r³ = 27 → r = 3.
  2. 2x = 3·3 = 9 and y = 9·3 = 27.

Final answer

9 and 27.

Step-by-step solution

  1. 1Set (aⁿ⁺¹ + bⁿ⁺¹)/(aⁿ + bⁿ) = √(ab).
  2. 2Cross-multiply and divide both sides by (ab)ⁿᐟ².
  3. 3This gives n = −1/2 (with the simplification aⁿ⁺¹ = a·aⁿ).

Final answer

n = −1/2.

Step-by-step solution

  1. 1Let the numbers be x and y: x + y = 6√(xy).
  2. 2Divide by y: t + 1 = 6√t where t = x/y.
  3. 3Let u = √t: u² − 6u + 1 = 0 → u = 3 ± 2√2.
  4. 4x/y = (3 + 2√2)² but (3+2√2)/(3−2√2) = (3+2√2)², so the ratio is (3 + 2√2) : (3 − 2√2).

Final answer

Ratio is (3 + 2√2) : (3 − 2√2).

Step-by-step solution

  1. 1Let the numbers be a and b: A = (a+b)/2, G = √(ab).
  2. 2a, b are the roots of x² − 2Ax + G² = 0.
  3. 3x = A ± √(A² − G²) = A ± √((A − G)(A + G)).
  4. 4Hence the numbers are A ± √((A+G)(A−G)).

Final answer

Proved: numbers are A ± √((A+G)(A−G)).

Step-by-step solution

  1. 1After n hours: 30·2ⁿ (G.P. with a = 30, r = 2).
  2. 22nd hour: 30·2² = 120; 4th hour: 30·2⁴ = 480.
  3. 3nth hour: 30·2ⁿ.

Final answer

120, 480, and 30·2ⁿ.

Step-by-step solution

  1. 1Amount = P(1 + r/100)ⁿ with P = 500, r = 10, n = 10.
  2. 2= 500(1.1)¹⁰.
  3. 3= 500 × 2.5937 ≈ Rs 1296.87.

Final answer

500(1.1)¹⁰ ≈ Rs 1296.87.

Step-by-step solution

  1. 1Sum of roots = 2 × A.M. = 16.
  2. 2Product of roots = G.M.² = 25.
  3. 3Equation: x² − 16x + 25 = 0.

Final answer

x² − 16x + 25 = 0.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Arithmetic progression

Geometric progression

Standard sums

Sum of cubes

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

  • The common difference or ratio can be negative, but every AP and GP sum formula still holds — do not reject a negative r.
  • AP-GP questions usually hide an arithmetic progression of exponents, so look for the index n in a pattern like 1, 3, 5 rather than treating each term separately.

FAQ

Frequently asked questions

How many questions are in NCERT Class 11 Maths Chapter 8 (Sequences and Series)?

There are 2 exercise questions in this chapter, numbered Ex 8.1 to Ex 8.2. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Sequences and Series Class 11 Maths?

The formulas this chapter's questions actually turn on are: Arithmetic progression, Geometric progression, Standard sums, Sum of cubes. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Sequences and Series important for JEE Main?

Very important — AP, GP and their sum formulas underpin progressions, binomial coefficients and calculus, and they are asked in every board and JEE Main paper.

Next Chapters

Continue learning

Previous: Binomial TheoremNext: Straight LinesSequences and Series revision notesAll Class 11 Maths NCERT Solutions

Same solutions, live doubt-clearing help

Reading a solution is step one — getting a doubt resolved in real time is what clears it. ClassApna runs small-batch CBSE, JEE & NEET coaching with daily doubt sessions and mock tests.

Book a Free Demo ClassExplore Plans & Batches

Small batches · 1-on-1 personal mentorship · Live online & offline centre