NCERT Solutions for Class 11 Maths Chapter 8: Sequences and Series (NCERT 2026–27)

These Class 11 Maths Chapter 8 solutions cover Sequences and Series completely. Every question of Exercise 8.1, Exercise 8.2 and the Miscellaneous Exercise on Chapter 8 is reproduced verbatim from the NCERT textbook and solved step by step — nth terms, geometric progressions, geometric mean, the A.M.–G.M. relationship and special series — with answers cross-checked against the book’s answer key.

Class: 11 Subject: Mathematics Chapter: 8 – Sequences and Series Exercises: 8.1, 8.2, Miscellaneous Session: 2026–27 Type: Solved NCERT exercises
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Chapter 8 Overview

Chapter 8, Sequences and Series, treats a sequence as a function whose domain is the set of natural numbers, and a series as the sum of the terms of a sequence. After revising arithmetic progressions, the chapter focuses on the geometric progression (G.P.) — its general term an = arn−1, the sum of n terms, the geometric mean of two numbers and the insertion of geometric means, the relationship A ≥ G between arithmetic and geometric means, and applications such as compound interest, depreciation and population growth. These Class 11 Maths Chapter 8 solutions solve every exercise and the Miscellaneous Exercise in full.

Key Concepts & Definitions

Sequence: an ordered list of numbers a1, a2, a3, … following a rule; the nth term an is the general term. A sequence with a fixed number of terms is finite, otherwise infinite.

Series: the indicated sum a1 + a2 + a3 + …, often written in sigma notation Σ.

Geometric progression (G.P.): a sequence in which each term (after the first) is a constant multiple of the preceding one. The constant r = ak+1/ak is the common ratio and a is the first term.

Geometric mean (G.M.): the G.M. of two positive numbers a and b is √(ab); then a, √(ab), b are in G.P.

A.M.–G.M. inequality: for two positive numbers, the arithmetic mean A = (a + b)/2 and the geometric mean G = √(ab) satisfy A ≥ G, with equality only when a = b.

Important Formulas (Chapter 8)

nth term of a G.P.: an = arn−1.

Sum of n terms of a G.P.: Sn = a(rn − 1)/(r − 1) = a(1 − rn)/(1 − r), for r ≠ 1; and Sn = na when r = 1.

Geometric mean: G = √(ab); for n means between a and b, r = (b/a)1/(n+1) and Gk = a(b/a)k/(n+1).

A.M. and G.M.: A = (a + b)/2, G = √(ab), and A ≥ G.

Special sums: Σk = n(n + 1)/2, Σk2 = n(n + 1)(2n + 1)/6, Σk3 = [n(n + 1)/2]2.

Exercise 8.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given at the back of the book.

Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

1. an = n(n + 2)

SOLUTION Put n = 1, 2, 3, 4, 5: a1 = 1·3 = 3, a2 = 2·4 = 8, a3 = 3·5 = 15, a4 = 4·6 = 24, a5 = 5·7 = 35. First five terms: 3, 8, 15, 24, 35.

2. an = n/(n + 1)

SOLUTION a1 = 1/2, a2 = 2/3, a3 = 3/4, a4 = 4/5, a5 = 5/6. First five terms: 1/2, 2/3, 3/4, 4/5, 5/6.

3. an = 2n

SOLUTION a1 = 2, a2 = 4, a3 = 8, a4 = 16, a5 = 32. First five terms: 2, 4, 8, 16, 32.

4. an = (2n − 3)/6

SOLUTION a1 = (2−3)/6 = −1/6, a2 = 1/6, a3 = 3/6 = 1/2, a4 = 5/6, a5 = 7/6. First five terms: −1/6, 1/6, 1/2, 5/6, 7/6.

5. an = (−1)n−1 5n+1

SOLUTION a1 = (+1)52 = 25, a2 = (−1)53 = −125, a3 = 54 = 625, a4 = −55 = −3125, a5 = 56 = 15625. First five terms: 25, −125, 625, −3125, 15625.

6. an = n(n2 + 5)/4

SOLUTION a1 = 1(6)/4 = 6/4 = 3/2, a2 = 2(9)/4 = 18/4 = 9/2, a3 = 3(14)/4 = 42/4 = 21/2, a4 = 4(21)/4 = 21, a5 = 5(30)/4 = 150/4 = 75/2. First five terms: 3/2, 9/2, 21/2, 21, 75/2.

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

7. an = 4n − 3; a17, a24

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

8. an = n2/2n; a7

SOLUTION a7 = 72/27 = 49/128. a7 = 49/128.

9. an = (−1)n−1 n3; a9

SOLUTION a9 = (−1)8(9)3 = (+1)(729) = 729.

10. an = n(n − 2)/(n + 3); a20

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

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:

11. a1 = 3, an = 3an−1 + 2 for all n > 1

SOLUTION a1 = 3; a2 = 3(3) + 2 = 11; a3 = 3(11) + 2 = 35; a4 = 3(35) + 2 = 107; a5 = 3(107) + 2 = 323. First five terms: 3, 11, 35, 107, 323. Series: 3 + 11 + 35 + 107 + 323 + …

12. a1 = −1, an = an−1/n, n ≥ 2

SOLUTION a1 = −1; a2 = a1/2 = −1/2; a3 = a2/3 = −1/6; a4 = a3/4 = −1/24; a5 = a4/5 = −1/120. First five terms: −1, −1/2, −1/6, −1/24, −1/120. Series: −1 + (−1/2) + (−1/6) + (−1/24) + (−1/120) + …

13. a1 = a2 = 2, an = an−1 − 1, n > 2

SOLUTION a1 = 2, a2 = 2; a3 = a2 − 1 = 1; a4 = a3 − 1 = 0; a5 = a4 − 1 = −1. First five terms: 2, 2, 1, 0, −1. Series: 2 + 2 + 1 + 0 + (−1) + …

14. The Fibonacci sequence is defined by 1 = a1 = a2 and an = an−1 + an−2, n > 2. Find an+1/an, for n = 1, 2, 3, 4, 5.

SOLUTION First compute terms: a1 = 1, a2 = 1, a3 = 2, a4 = 3, a5 = 5, a6 = 8. n = 1: a2/a1 = 1/1 = 1. n = 2: a3/a2 = 2/1 = 2. n = 3: a4/a3 = 3/2 = 3/2. n = 4: a5/a4 = 5/3 = 5/3. n = 5: a6/a5 = 8/5 = 8/5.

Exercise 8.2 Solutions

1. Find the 20th and nth terms of the G.P. 5/2, 5/4, 5/8, …

SOLUTION Here a = 5/2 and r = (5/4)/(5/2) = 1/2. an = arn−1 = (5/2)(1/2)n−1 = 5/2n. a20 = 5/220. ∴ nth term = 5/2n and 20th term = 5/220.

2. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.

SOLUTION a8 = ar7 = 192 with r = 2, so a·128 = 192, giving a = 3/2. a12 = ar11 = (3/2)(2)11 = (3/2)(2048) = 3072. (Equivalently a12 = a8·r4 = 192 × 16 = 3072.)

3. The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.

SOLUTION p = ar4, q = ar7, s = ar10. q2 = (ar7)2 = a2r14. ps = (ar4)(ar10) = a2r14. ∴ q2 = ps. (Hence p, q, s are themselves in G.P.) ✓

4. The 4th term of a G.P. is square of its second term, and the first term is −3. Determine its 7th term.

SOLUTION a = −3. Given a4 = (a2)2: ar3 = (ar)2 = a2r2. Divide by ar2 (≠ 0): r = a = −3. a7 = ar6 = (−3)(−3)6 = (−3)(729) = −2187.

5. Which term of the following sequences: (a) 2, 2√2, 4, … is 128? (b) √3, 3, 3√3, … is 729? (c) 1/3, 1/9, 1/27, … is 1/19683?

SOLUTION (a) a = 2, r = √2. an = 2(√2)n−1 = 21+(n−1)/2. Set = 128 = 27: 1 + (n−1)/2 = 7 ⇒ (n−1)/2 = 6 ⇒ n = 13th term. (b) a = √3 = 31/2, r = √3. an = 3n/2. Set = 729 = 36: n/2 = 6 ⇒ n = 12th term. (c) a = 1/3, r = 1/3. an = (1/3)n. Set = 1/19683 = (1/3)9 (since 39 = 19683): n = 9th term.

6. For what values of x, the numbers −2/7, x, −7/2 are in G.P.?

SOLUTION For a G.P. the middle term satisfies x2 = (−2/7)(−7/2) = 1. x = ±1. ∴ x = ±1.

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10:

7. 0.15, 0.015, 0.0015, … 20 terms.

SOLUTION a = 0.15, r = 0.1, n = 20. S20 = a(1 − rn)/(1 − r) = 0.15(1 − (0.1)20)/(1 − 0.1) = 0.15(1 − (0.1)20)/0.9. S20 = (1/6)[1 − (0.1)20].

8. √7, √21, 3√7, … n terms.

SOLUTION a = √7, r = √21/√7 = √3. Sn = a(rn − 1)/(r − 1) = √7((√3)n − 1)/(√3 − 1). Rationalising by multiplying numerator and denominator by (√3 + 1): Sn = [√7(√3 + 1)/2]·(3n/2 − 1). Sn = (√7(√3 + 1)/2)(3n/2 − 1).

9. 1, −a, a2, −a3, … n terms (if a ≠ −1).

SOLUTION First term = 1, common ratio r = −a. Sn = (1 − rn)/(1 − r) = (1 − (−a)n)/(1 + a). Sn = (1 − (−a)n)/(1 + a).

10. x3, x5, x7, … n terms (if x ≠ ±1).

SOLUTION a = x3, r = x2. Sn = a(rn − 1)/(r − 1) = x3((x2)n − 1)/(x2 − 1) = x3(x2n − 1)/(x2 − 1). Sn = x3(1 − x2n)/(1 − x2).

11. Evaluate Σk=111 (2 + 3k).

SOLUTION Σk=111(2 + 3k) = Σk=1112 + Σk=1113k. Σ2 = 2 × 11 = 22. Σ3k = 3 + 32 + … + 311 = 3(311 − 1)/(3 − 1) = (3/2)(311 − 1). Total = 22 + (3/2)(311 − 1) = 22 + (3/2)(311 − 1).

12. The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.

SOLUTION Let the terms be a/r, a, ar. Product = a3 = 1 ⇒ a = 1. Sum: 1/r + 1 + r = 39/10 ⇒ 1/r + r = 29/10 ⇒ 10r2 − 29r + 10 = 0. Solve: r = [29 ± √(841 − 400)]/20 = [29 ± 21]/20 = 5/2 or 2/5. For r = 5/2: terms are 2/5, 1, 5/2. For r = 2/5: terms are 5/2, 1, 2/5.

13. How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?

SOLUTION a = 3, r = 3. Sn = 3(3n − 1)/(3 − 1) = (3/2)(3n − 1) = 120. 3n − 1 = 80 ⇒ 3n = 81 = 34 ⇒ n = 4. ∴ 4 terms are needed.

14. The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.

SOLUTION Sum of first 3 terms = a(r3 − 1)/(r − 1) = 16. Sum of next 3 terms (4th to 6th) = r3·[a(r3 − 1)/(r − 1)] = 128. Dividing: r3 = 128/16 = 8 ⇒ r = 2. Then a(23 − 1)/(2 − 1) = 7a = 16 ⇒ a = 16/7. Sn = a(rn − 1)/(r − 1) = (16/7)(2n − 1). ∴ a = 16/7, r = 2, Sn = (16/7)(2n − 1).

15. Given a G.P. with a = 729 and 7th term 64, determine S7.

SOLUTION a7 = ar6 = 64 ⇒ 729 r6 = 64 ⇒ r6 = 64/729 = (2/3)6 ⇒ r = ±2/3. For r = 2/3: S7 = a(1 − r7)/(1 − r) = 729(1 − (2/3)7)/(1/3) = 3 × 729 × (1 − 128/2187) = 2187 × (2059/2187) = 2059. For r = −2/3: S7 = 729(1 − (−2/3)7)/(1 + 2/3) = 729(1 + 128/2187)/(5/3) = (3/5)(729)(2315/2187) = 463. ∴ S7 = 2059 or 463.

16. Find a G.P. for which sum of the first two terms is −4 and the fifth term is 4 times the third term.

SOLUTION 5th term = 4 × 3rd term: ar4 = 4ar2 ⇒ r2 = 4 ⇒ r = ±2. Sum of first two terms: a + ar = −4 ⇒ a(1 + r) = −4. For r = 2: a(3) = −4 ⇒ a = −4/3. G.P.: −4/3, −8/3, −16/3, … For r = −2: a(−1) = −4 ⇒ a = 4. G.P.: 4, −8, 16, −32, 64, …

17. If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.

SOLUTION x = ar3, y = ar9, z = ar15. y2 = (ar9)2 = a2r18, and xz = (ar3)(ar15) = a2r18. So y2 = xz, which means x, y, z are in G.P. ✓

18. Find the sum to n terms of the sequence, 8, 88, 888, 8888…

SOLUTION Sn = 8 + 88 + 888 + … = (8/9)(9 + 99 + 999 + … to n terms). = (8/9)[(10 − 1) + (102 − 1) + … + (10n − 1)] = (8/9)[(10 + 102 + … + 10n) − n]. = (8/9)[10(10n − 1)/9 − n]. Sn = (80/81)(10n − 1) − (8/9)n.

19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2.

SOLUTION Products: (2×128), (4×32), (8×8), (16×2), (32×1/2) = 256, 128, 64, 32, 16. This is a G.P. with a = 256, r = 1/2, n = 5. Sum = 256(1 − (1/2)5)/(1 − 1/2) = 256 × 2 × (31/32) = 512 × 31/32 = 496.

20. Show that the products of the corresponding terms of the sequences a, ar, ar2, … arn−1 and A, AR, AR2, … ARn−1 form a G.P, and find the common ratio.

SOLUTION The product sequence is aA, (ar)(AR), (ar2)(AR2), … i.e. aA, aA(rR), aA(rR)2, … Each term is (rR) times the previous one, so it is a G.P. with first term aA. Common ratio = rR. ✓

21. Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.

SOLUTION Let the terms be a, ar, ar2, ar3. Given ar2 − a = 9 and ar − ar3 = 18. a(r2 − 1) = 9 …(i); ar(1 − r2) = 18 …(ii). Divide (ii) by (i): [ar(1 − r2)]/[a(r2 − 1)] = 18/9 ⇒ −r = 2 ⇒ r = −2. From (i): a(4 − 1) = 9 ⇒ a = 3. Terms: 3, −6, 12, −24.

22. If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that aq−r br−p cp−q = 1.

SOLUTION Let first term A and ratio R. Then a = ARp−1, b = ARq−1, c = ARr−1. aq−r br−p cp−q = A(q−r)+(r−p)+(p−q) × R(p−1)(q−r)+(q−1)(r−p)+(r−1)(p−q). Exponent of A = (q−r)+(r−p)+(p−q) = 0. Exponent of R = [p(q−r)+q(r−p)+r(p−q)] − [(q−r)+(r−p)+(p−q)] = 0 − 0 = 0. ∴ aq−r br−p cp−q = A0R0 = 1. ✓

23. If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab)n.

SOLUTION Terms: a, ar, ar2, …, arn−1; here b = arn−1. P = an r0+1+…+(n−1) = an rn(n−1)/2. P2 = a2n rn(n−1). (ab)n = (a · arn−1)n = (a2rn−1)n = a2n rn(n−1). ∴ P2 = (ab)n. ✓

24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is 1/rn.

SOLUTION Sum of first n terms S1 = a(rn − 1)/(r − 1). The terms from (n+1)th to (2n)th have first term arn, so their sum S2 = arn(rn − 1)/(r − 1) = rn·S1. ∴ S1/S2 = S1/(rnS1) = 1/rn. ✓

25. If a, b, c and d are in G.P. show that (a2 + b2 + c2)(b2 + c2 + d2) = (ab + bc + cd)2.

SOLUTION Since a, b, c, d are in G.P., b2 = ac, c2 = bd and bc = ad. R.H.S. = (ab + bc + cd)2. Replace bc = ad in two ways and expand: (ab + bc + cd)2 = a2b2 + b2c2 + c2d2 + 2(ab·bc + bc·cd + ab·cd). L.H.S. = (a2+b2+c2)(b2+c2+d2). On expanding and using b2=ac, c2=bd, bc=ad, the cross terms 2ab2c + 2bc·cd + 2ab·cd appear, matching R.H.S. exactly. Hence L.H.S. = R.H.S. ✓

26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

SOLUTION Let 3, G1, G2, 81 be a G.P. with 81 as the 4th term: 81 = 3r3 ⇒ r3 = 27 ⇒ r = 3. G1 = 3×3 = 9, G2 = 3×32 = 27. ∴ the numbers are 9 and 27.

27. Find the value of n so that (an+1 + bn+1)/(an + bn) may be the geometric mean between a and b.

SOLUTION Require (an+1 + bn+1)/(an + bn) = √(ab) = a1/2b1/2. Cross-multiply: an+1 + bn+1 = a1/2b1/2(an + bn) = an+1/2b1/2 + a1/2bn+1/2. an+1 − an+1/2b1/2 = a1/2bn+1/2 − bn+1 ⇒ an+1/2(a1/2 − b1/2) = bn+1/2(a1/2 − b1/2). So an+1/2 = bn+1/2 ⇒ (a/b)n+1/2 = 1 ⇒ n + 1/2 = 0 ⇒ n = −1/2.

28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3 + 2√2) : (3 − 2√2).

SOLUTION Let the numbers be a and b. Given a + b = 6√(ab). Using componendo-dividendo with (a + b) and 2√(ab): (a + b + 2√(ab))/(a + b − 2√(ab)) = (6√(ab) + 2√(ab))/(6√(ab) − 2√(ab)) = 8/4 = 2. So (√a + √b)2/(√a − √b)2 = 2 ⇒ (√a + √b)/(√a − √b) = √2. Again by componendo-dividendo: √a/√b = (√2 + 1)/(√2 − 1). Squaring: a/b = (√2 + 1)2/(√2 − 1)2 = (3 + 2√2)/(3 − 2√2). ∴ a : b = (3 + 2√2) : (3 − 2√2). ✓

29. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A ± √((A + G)(A − G)).

SOLUTION Let the numbers be a and b. A = (a + b)/2 and G = √(ab), so a + b = 2A and ab = G2. (a − b)2 = (a + b)2 − 4ab = 4A2 − 4G2 = 4(A + G)(A − G). a − b = 2√((A + G)(A − G)). Adding/subtracting with a + b = 2A: a, b = A ± √((A + G)(A − G)). ✓

30. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour?

SOLUTION Population forms a G.P. with a = 30 and r = 2; after t hours it is 30(2)t. End of 2nd hour: 30(2)2 = 120. End of 4th hour: 30(2)4 = 480. End of nth hour: 30(2)n.

31. What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

SOLUTION Amount after each year forms a G.P. with ratio (1 + 10/100) = 1.1. After 10 years: A = 500(1.1)10. ∴ the amount is Rs 500(1.1)10 (≈ Rs 1296.87).

32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.

SOLUTION Let roots be α, β. A.M. = (α + β)/2 = 8 ⇒ α + β = 16. G.M. = √(αβ) = 5 ⇒ αβ = 25. Equation: x2 − (sum)x + (product) = 0. ∴ x2 − 16x + 25 = 0.

Miscellaneous Exercise on Chapter 8 Solutions

1. If f is a function satisfying f(x + y) = f(x)f(y) for all x, y ∈ N such that f(1) = 3 and Σx=1n f(x) = 120, find the value of n.

SOLUTION f(1) = 3, f(2) = f(1)f(1) = 9, f(3) = f(1)f(2) = 27, … so f(x) = 3x, a G.P. with a = 3, r = 3. Σx=1n f(x) = 3(3n − 1)/(3 − 1) = (3/2)(3n − 1) = 120. 3n − 1 = 80 ⇒ 3n = 81 = 34 ⇒ n = 4.

2. The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

SOLUTION Sn = a(rn − 1)/(r − 1) = 5(2n − 1)/(2 − 1) = 5(2n − 1) = 315. 2n − 1 = 63 ⇒ 2n = 64 = 26 ⇒ n = 6. Last term a6 = ar5 = 5(2)5 = 5 × 32 = 160. ∴ last term = 160, number of terms = 6.

3. The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.

SOLUTION a = 1. Third term = r2, fifth term = r4; r2 + r4 = 90. Let r2 = t: t2 + t − 90 = 0 ⇒ (t + 10)(t − 9) = 0 ⇒ t = 9 (reject t = −10). r2 = 9 ⇒ r = ±3. ∴ common ratio = ±3.

4. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

SOLUTION Let the numbers be a, ar, ar2. Sum: a(1 + r + r2) = 56 …(i). (a − 1), (ar − 7), (ar2 − 21) are in A.P.: 2(ar − 7) = (a − 1) + (ar2 − 21). 2ar − 14 = a + ar2 − 22 ⇒ a(r2 − 2r + 1) = 8 ⇒ a(r − 1)2 = 8 …(ii). From (i), a = 56/(1 + r + r2). Substitute in (ii): 56(r − 1)2 = 8(1 + r + r2) ⇒ 7(r2 − 2r + 1) = 1 + r + r2 ⇒ 6r2 − 15r + 6 = 0 ⇒ 2r2 − 5r + 2 = 0 ⇒ r = 2 or 1/2. For r = 2: a = 56/7 = 8 ⇒ numbers 8, 16, 32. For r = 1/2: a = 32 ⇒ numbers 32, 16, 8.

5. A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

SOLUTION Let there be 2n terms a, ar, …, ar2n−1. Sum of all terms S = a(r2n − 1)/(r − 1). Odd-place terms a, ar2, ar4, … (n terms, ratio r2): Sodd = a((r2)n − 1)/(r2 − 1) = a(r2n − 1)/(r2 − 1). Given S = 5 Sodd: a(r2n − 1)/(r − 1) = 5a(r2n − 1)/(r2 − 1). Cancel a(r2n − 1): 1/(r − 1) = 5/(r2 − 1) = 5/[(r − 1)(r + 1)] ⇒ r + 1 = 5 ⇒ r = 4.

6. If (a + bx)/(a − bx) = (b + cx)/(b − cx) = (c + dx)/(c − dx) (x ≠ 0), then show that a, b, c and d are in G.P.

SOLUTION From (a + bx)/(a − bx) = (b + cx)/(b − cx), apply componendo-dividendo: 2a/2bx = 2b/2cx ⇒ a/(bx) = b/(cx) ⇒ b2 = ac. Similarly, from the second and third ratios: c2 = bd. b2 = ac and c2 = bd ⇒ b/a = c/b = d/c, so a, b, c, d are in G.P. ✓

7. Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P2Rn = Sn.

SOLUTION Terms a, ar, …, arn−1. S = a(rn − 1)/(r − 1). P = an rn(n−1)/2. R = (1/a)(1 − (1/r)n)/(1 − 1/r) = (rn − 1)/(a rn−1(r − 1)). S/R = [a(rn−1)/(r−1)] ÷ [(rn−1)/(a rn−1(r−1))] = a2 rn−1. So (S/R)n = a2n rn(n−1) = (anrn(n−1)/2)2 = P2. ∴ Sn = P2Rn. ✓

8. If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) are in G.P.

SOLUTION Let common ratio be r, so b = ar, c = ar2, d = ar3. an + bn = an(1 + rn); bn + cn = anrn(1 + rn); cn + dn = anr2n(1 + rn). Each is rn times the previous one, so they form a G.P. with ratio rn. Equivalently (bn+cn)2 = (an+bn)(cn+dn). ✓

9. If a and b are the roots of x2 − 3x + p = 0 and c, d are roots of x2 − 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q − p) = 17 : 15.

SOLUTION a + b = 3, ab = p; c + d = 12, cd = q. Let a, b, c, d be in G.P. with ratio r: b = ar, c = ar2, d = ar3. a + ar = 3 ⇒ a(1 + r) = 3; ar2 + ar3 = 12 ⇒ ar2(1 + r) = 12. Dividing: r2 = 4 ⇒ r = 2 (positive root). Then a(3) = 3 ⇒ a = 1. So a = 1, b = 2, c = 4, d = 8. p = ab = 2, q = cd = 32. (q + p)/(q − p) = (32 + 2)/(32 − 2) = 34/30 = 17/15. ∴ (q + p) : (q − p) = 17 : 15. ✓

10. The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show that a : b = (m + √(m2 − n2)) : (m − √(m2 − n2)).

SOLUTION A.M./G.M. = m/n ⇒ ((a + b)/2)/√(ab) = m/n ⇒ (a + b)/(2√(ab)) = m/n. By componendo-dividendo: (a + b + 2√(ab))/(a + b − 2√(ab)) = (m + n)/(m − n) ⇒ (√a + √b)2/(√a − √b)2 = (m + n)/(m − n). √a/√b = (√(m+n) + √(m−n))/(√(m+n) − √(m−n)). Squaring: a/b = ((m + n) + 2√(m2−n2) + (m − n))/((m + n) − 2√(m2−n2) + (m − n)). a/b = (2m + 2√(m2−n2))/(2m − 2√(m2−n2)) = (m + √(m2−n2))/(m − √(m2−n2)). ✓

11. Find the sum of the following series up to n terms: (i) 5 + 55 + 555 + … (ii) .6 + .66 + .666 + …

SOLUTION (i) Sn = 5 + 55 + 555 + … = (5/9)(9 + 99 + 999 + …) = (5/9)[(10 − 1) + (102 − 1) + …] = (5/9)[10(10n − 1)/9 − n]. (i) Sn = (50/81)(10n − 1) − (5/9)n. (ii) Sn = .6 + .66 + .666 + … = (6/9)[(1 − 0.1) + (1 − 0.01) + … to n terms] = (2/3)[n − (0.1 + 0.01 + … n terms)]. The bracketed G.P.: 0.1(1 − 0.1n)/(1 − 0.1) = (1/9)(1 − 10−n). So Sn = (2/3)[n − (1/9)(1 − 10−n)]. (ii) Sn = (2n/3) − (2/27)(1 − 10−n).

12. Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + … + n terms.

SOLUTION The nth term: factors are 2n and (2n + 2), so an = 2n(2n + 2) = 4n(n + 1) = 4n2 + 4n. a20 = 4(20)2 + 4(20) = 4(400) + 80 = 1600 + 80 = 1680.

13. A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?

SOLUTION Unpaid balance = 12000 − 6000 = Rs 6000, repaid in 12 instalments of Rs 500. Interest (12%) on outstanding amounts 6000, 5500, …, 500 is paid each year. Total interest = 12% of (6000 + 5500 + … + 500). Sum = 500 + 1000 + … + 6000 = (12/2)(500 + 6000) = 6 × 6500 = 39000. Interest = 12% of 39000 = 4680. Total cost = 12000 + 4680 = Rs 16680.

14. Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

SOLUTION Unpaid balance = 22000 − 4000 = Rs 18000, repaid in 18 instalments of Rs 1000. Total interest = 10% of (18000 + 17000 + … + 1000). Sum = 1000 + 2000 + … + 18000 = (18/2)(1000 + 18000) = 9 × 19000 = 171000. Interest = 10% of 171000 = 17100. Total cost = 22000 + 17100 = Rs 39100.

15. A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.

SOLUTION Number of letters mailed in successive sets: 4, 42, 43, …, 48 — a G.P. with a = 4, r = 4. Total letters = 4(48 − 1)/(4 − 1) = (4/3)(65536 − 1) = (4/3)(65535) = 87380. Cost = 87380 × Rs 0.50 = Rs 43690.

16. A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.

SOLUTION Simple interest each year = 5% of 10000 = Rs 500. Amount in 15th year = principal + interest for 14 completed years = 10000 + 14 × 500 = 10000 + 7000 = Rs 17000. Total amount after 20 years = 10000 + 20 × 500 = 10000 + 10000 = Rs 20000.

17. A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.

SOLUTION Each year the value becomes (1 − 20/100) = 4/5 of the previous value — a G.P. with ratio 4/5. Value after 5 years = 15625(4/5)5 = 15625 × 1024/3125 = 5 × 1024 = Rs 5120.

18. 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.

SOLUTION If the work was originally planned for x days, total planned work = 150x worker-days. Actually it took (x + 8) days with workers 150, 146, 142, … (an A.P., first term 150, d = −4, (x + 8) terms). Work done = ((x + 8)/2)[2(150) + (x + 8 − 1)(−4)] = ((x + 8)/2)[300 − 4(x + 7)] = ((x + 8)/2)(272 − 4x). Equate to 150x: ((x + 8)/2)(272 − 4x) = 150x ⇒ (x + 8)(136 − 2x) = 150x. 136x − 2x2 + 1088 − 16x = 150x ⇒ −2x2 − 30x + 1088 = 0 ⇒ x2 + 15x − 544 = 0 ⇒ (x − 17)(x + 32) = 0 ⇒ x = 17. Work was completed in x + 8 = 17 + 8 = 25 days.

Common Mistakes to Avoid

Watch out for these

  • Using the A.P. formula a + (n − 1)d instead of the G.P. formula arn−1 for the nth term — check whether the ratio or the difference is constant first.
  • Forgetting that r can be negative (e.g. Q15, Q3 of Miscellaneous) — always give both values when r2 or r6 is found.
  • Dropping the “1 − rn” (or rn − 1) factor or writing the sum formula with the wrong sign of (r − 1).
  • For “7 + 77 + 777” type series, forgetting to take out the factor (digit/9) and subtract the n ones.
  • Confusing simple interest (linear, A.P.) with compound interest/depreciation (geometric, G.P.).
  • In A.M.–G.M. ratio problems, forgetting to use componendo-dividendo — squaring directly leads to messy algebra.

Practice MCQs & Assertion–Reason

1. The common ratio of the G.P. 5/2, 5/4, 5/8, … is:

(a) 2    (b) 1/2    (c) 5/2    (d) 5

2. If the 8th term of a G.P. with r = 2 is 192, its first term is:

(a) 3    (b) 3/2    (c) 6    (d) 1.5 × 27

3. The geometric mean of 4 and 16 is:

(a) 6    (b) 8    (c) 10    (d) 12

4. For two positive numbers, which is always true?

(a) A < G    (b) A = G    (c) A ≥ G    (d) A ≤ G

5. Which term of the G.P. 2, 2√2, 4, … is 128?

(a) 11th    (b) 12th    (c) 13th    (d) 14th

6. The number of terms of the G.P. 3, 32, 33, … whose sum is 120 is:

(a) 3    (b) 4    (c) 5    (d) 6

7. If x, y, z are the 4th, 10th and 16th terms of a G.P., then:

(a) y2 = x + z    (b) y2 = xz    (c) y = xz    (d) 2y = x + z

8. The value of x for which −2/7, x, −7/2 are in G.P. is:

(a) 1 only    (b) −1 only    (c) ±1    (d) ±7/2

9. The quadratic equation whose roots have A.M. 8 and G.M. 5 is:

(a) x2 − 16x + 25 = 0    (b) x2 + 16x + 25 = 0    (c) x2 − 10x + 64 = 0    (d) x2 − 25x + 16 = 0

10. If a value depreciates 20% each year from Rs 15625, after 5 years it becomes:

(a) Rs 6250    (b) Rs 5120    (c) Rs 4096    (d) Rs 3125

Answer key: 1-(b), 2-(b), 3-(b), 4-(c), 5-(c), 6-(b), 7-(b), 8-(c), 9-(a), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The nth term of the G.P. 5/2, 5/4, 5/8, … is 5/2n.

Reason: In a G.P. the nth term is an = arn−1.

A-R 2. Assertion: If the 5th, 8th and 11th terms of a G.P. are p, q, s, then q2 = ps.

Reason: Any three consecutive terms of a G.P. satisfy (middle)2 = (first)(third), and p, q, s are equally spaced terms.

A-R 3. Assertion: For two positive numbers, A.M. is always less than G.M.

Reason: A − G = (√a − √b)2/2 ≥ 0.

A-R 4. Assertion: The sum 3 + 32 + 33 + … to 4 terms equals 120.

Reason: The sum of n terms of a G.P. is a(rn − 1)/(r − 1) for r ≠ 1.

A-R 5. Assertion: To make −2/7, x, −7/2 a G.P., x must equal 1.

Reason: The geometric mean condition gives x2 = (−2/7)(−7/2) = 1.

Answer key: 1-(A), 2-(A), 3-(D), 4-(B), 5-(D).

Quick Revision Summary

  • A sequence is a function on natural numbers; a series is the sum of its terms (sigma notation Σ).
  • G.P.: each term = previous term × r (common ratio); nth term an = arn−1.
  • Sum of n terms: Sn = a(rn − 1)/(r − 1) = a(1 − rn)/(1 − r), r ≠ 1; Sn = na when r = 1.
  • Geometric mean of a and b is √(ab); to insert n means, r = (b/a)1/(n+1).
  • For positive numbers, A.M. ≥ G.M., with equality only when the numbers are equal.
  • Series like 7 + 77 + 777 + … are handled by factoring out (digit/9) and relating to a G.P. of powers of 10.
  • Compound interest, depreciation and population growth are G.P. models; simple interest is an A.P. model.

How to score full marks in this chapter

Always identify a and r first, then choose the correct sum formula and the correct sign of (r − 1). Show the substitution explicitly so each step earns its mark. For “which term” problems, write the nth term as a single power and compare exponents. In word problems, decide whether the model is geometric (compound interest, doubling, depreciation) or arithmetic (simple interest, dropping workers) before you start — this single decision usually carries the whole answer.

Frequently Asked Questions

What is Class 11 Maths Chapter 8 about?

Chapter 8, Sequences and Series, covers sequences and series, geometric progressions (general term and sum of n terms), the geometric mean, inserting geometric means, the relationship A ≥ G between arithmetic and geometric means, and applications such as compound interest, depreciation and population growth.

How many exercises are there in Class 11 Maths Chapter 8?

There are two numbered exercises — Exercise 8.1 (14 questions) and Exercise 8.2 (32 questions) — plus the Miscellaneous Exercise on Chapter 8 (18 questions). Every question is solved step by step on this page.

What is the formula for the sum of n terms of a G.P.?

For common ratio r ≠ 1, the sum of the first n terms is Sn = a(rn − 1)/(r − 1), which can also be written as a(1 − rn)/(1 − r). When r = 1, all terms are equal and Sn = na.

Are these Class 11 Maths Chapter 8 solutions free?

Yes. All solutions are free and follow the official NCERT Class 11 Mathematics textbook for the 2026–27 session, with answers verified against the book’s answer key.

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