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NCERT Solutions · Class 11 Maths · Chapter 8

Chapter 8: Sequences and Series (Algebra)

Q: Write the first five terms: (−1)ⁿ⁻¹5ⁿ⁺¹

25, −125, 625, −3125, 15625

Step-by-step solutions to all exercises of Chapter 8, Sequences and Series (NCERT Class 11 Maths, 2026-27): terms of sequences, arithmetic and geometric progressions, sums of GPs, AM–GM, and the Miscellaneous Exercise. All 64 questions are answered, with the key answer highlighted.

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GP: $a_{n} = a r^{n - 1}$ , $S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$ , $S_{\infty} = \frac{a}{1 - r}$ for |r| < 1. AP: $a_{n} = a + (n - 1) d$ , $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ . $A \geq G$ for positive numbers.

Exercise 8.1

Write the first five terms: aₙ = n(n + 2)

Solution

3, 8, 15, 24, 35.

3, 8, 15, 24, 35

Write the first five terms: n/(n + 1)

Solution

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

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

Write the first five terms: 2ⁿ

Solution

2, 4, 8, 16, 32.

2, 4, 8, 16, 32

Write the first five terms: (2n − 3)/6

Solution

−1/6, 1/6, 1/2, 5/6, 7/6.

−1/6, 1/6, 1/2, 5/6, 7/6

Write the first five terms: (−1)ⁿ⁻¹5ⁿ⁺¹

Solution

5², −5³, …: 25, −125, 625, −3125, 15625.

25, −125, 625, −3125, 15625

Write the first five terms: n(n² + 5)/4.

Solution

1(6)/4, 2(9)/4, 3(14)/4, 4(21)/4, 5(30)/4: 3/2, 9/2, 21/2, 21, 75/2.

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

Find: a₁₇, a₂₄ for aₙ = 4n − 3

Solution

68 − 3 = 65; 96 − 3 = 93.

65; 93

Find: a₇ for n²/2ⁿ

Solution

49/128.

See the working above.

Find: a₉ for (−1)ⁿ⁻¹n³

Solution

(−1)⁸(729) = 729.

729

Find: a₂₀ for n(n − 2)/(n + 3).

Solution

20 × 18/23 = 360/23.

360/23

First five terms and series: a₁ = 3, aₙ = 3aₙ₋₁ + 2

Solution

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

3 + 11 + 35 + 107 + 323 + …

First five terms and series: a₁ = −1, aₙ = aₙ₋₁/n

Solution

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

−1 − 1/2 − 1/6 − 1/24 − 1/120 − …

First five terms and series: a₁ = a₂ = 2, aₙ = aₙ₋₁ − 1 (n > 2).

Solution

2, 2, 1, 0, −1; 2 + 2 + 1 + 0 + (−1) + ….

2 + 2 + 1 + 0 + (−1) + …

Fibonacci: 1 = a₁ = a₂, aₙ = aₙ₋₁ + aₙ₋₂. Find aₙ₊₁/aₙ for n = 1, …, 5.

Solution

Terms 1, 1, 2, 3, 5, 8: ratios 1, 2, 3/2, 5/3, 8/5.

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

Exercise 8.2

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

Solution

a = 5/2, r = ½: a₂₀ = (5/2)(½)¹⁹ = 5/2²⁰; aₙ = 5/2ⁿ.

5/2²⁰; 5/2ⁿ

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

Solution

ar⁷ = 192 ⇒ a = 192/128 = 3/2; a₁₂ = (3/2)2¹¹ = 3072.

3072

The 5th, 8th, 11th terms are p, q, s. Show q² = ps.

Solution

p = ar⁴, q = ar⁷, s = ar¹⁰: ps = a²r¹⁴ = (ar⁷)² = q².

q²

The 4th term is the square of the 2nd and the first term is −3. Find the 7th term.

Solution

ar³ = (ar)² ⇒ r = a = −3; a₇ = ar⁶ = (−3)(729) = −2187.

−2187

Which term of (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) 2(√2)ⁿ⁻¹ = 128 ⇒ (√2)ⁿ⁻¹ = 64 = (√2)¹² ⇒ 13th. (b) (√3)ⁿ = 729 = (√3)¹² ⇒ 12th. (c) (1/3)ⁿ = 3⁻⁹ ⇒ 9th.

(a) 13th; (b) 12th; (c) 9th

For what x are −2/7, x, −7/2 in G.P.?

Solution

x² = (−2/7)(−7/2) = 1 ⇒ x = ±1.

x = ±1

Sum: 0.15, 0.015, 0.0015, … 20 terms

Solution

a = 0.15, r = 0.1: 0.15(1 − 0.1²⁰)/0.9 = (1/6)(1 − 0.1²⁰).

(1/6)(1 − 0.1²⁰)

Sum: √7, √21, 3√7, … n terms

Solution

r = √3: √7((√3)ⁿ − 1)/(√3 − 1) = (√7/2)(√3 + 1)(3^(n/2) − 1).

(√7/2)(√3 + 1)(3^(n/2) − 1)

Sum: 1, −a, a², −a³, … n terms (a ≠ −1)

Solution

r = −a: [1 − (−a)ⁿ]/(1 + a).

[1 − (−a)ⁿ]/(1 + a)

Sum: x³, x⁵, x⁷, … n terms (x ≠ ±1).

Solution

r = x²: x³(1 − x²ⁿ)/(1 − x²).

x³(1 − x²ⁿ)/(1 − x²)

Evaluate Σ_(k=1)¹¹(2 + 3ᵏ).

Solution

= 2 × 11 + (3 + 3² + … + 3¹¹) = 22 + 3(3¹¹ − 1)/2 = 22 + 265719 = 265741.

265741

The sum of the first three terms is 39/10 and their product 1. Find r and the terms.

Solution

a/r, a, ar: a³ = 1 ⇒ a = 1. 1/r + 1 + r = 39/10 ⇒ 10r² − 29r + 10 = 0 ⇒ r = 5/2 or 2/5: terms 2/5, 1, 5/2 (or 5/2, 1, 2/5).

r = 5/2 or 2/5; 2/5, 1, 5/2

How many terms of 3, 3², 3³, … give the sum 120?

Solution

3(3ⁿ − 1)/2 = 120 ⇒ 3ⁿ = 81 ⇒ n = 4.

n = 4

The sum of the first three terms is 16 and of the next three 128. Find a, r and Sₙ.

Solution

a(1 + r + r²) = 16; ar³(1 + r + r²) = 128 ⇒ r³ = 8 ⇒ r = 2, a = 16/7. Sₙ = (16/7)(2ⁿ − 1).

r = 2; Sₙ = (16/7)(2ⁿ − 1)

a = 729, a₇ = 64. Find S₇.

Solution

r⁶ = 64/729 ⇒ r = ±2/3. r = 2/3: S₇ = 729(1 − 128/2187)/(1/3) = 2187 − 128 = 2059; r = −2/3: 729(1 + 128/2187)/(5/3) = (2187 + 128)/5 = 463.

2059; 463

Find a G.P. with the sum of the first two terms −4 and the 5th term 4 times the 3rd.

Solution

ar⁴ = 4ar² ⇒ r = ±2. r = 2: 3a = −4, a = −4/3: −4/3, −8/3, −16/3, …; r = −2: −a = −4, a = 4: 4, −8, 16, −32, ….

−4/3, −8/3, −16/3, …; 4, −8, 16, −32, …

The 4th, 10th, 16th terms are x, y, z. Prove x, y, z are in G.P.

Solution

x = ar³, y = ar⁹, z = ar¹⁵: xz = a²r¹⁸ = y²: in G.P.

in G.P.

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

Solution

(8/9)(9 + 99 + 999 + …) = (8/9)[10(10ⁿ − 1)/9 − n] = (80/81)(10ⁿ − 1) − 8n/9.

(80/81)(10ⁿ − 1) − 8n/9

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

Solution

256 + 128 + 64 + 32 + 16 = 496 (a G.P. with a = 256, r = ½).

496

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

Solution

Products aA, aA(rR), aA(rR)², …: each is rR times the previous: a G.P. with ratio rR.

ratio rR

Find four numbers in G.P. whose 3rd term exceeds the 1st by 9 and the 2nd exceeds the 4th by 18.

Solution

a, ar, ar², ar³: a(r² − 1) = 9, ar(1 − r²) = 18 ⇒ −r = 2 ⇒ r = −2, a = 3: 3, −6, 12, −24.

3, −6, 12, −24

The pth, qth, rth terms are a, b, c. Prove a^(q−r)bʳ⁻ᵖc^(p−q) = 1.

Solution

a = ARᵖ⁻¹, b = AR^(q−1), c = ARʳ⁻¹. The power of A: (q − r) + (r − p) + (p − q) = 0; of R: (p − 1)(q − r) + (q − 1)(r − p) + (r − 1)(p − q) = 0 (expand). So the product = A⁰R⁰ = 1.

The first and nth terms are a and b; P is the product of n terms. Prove P² = (ab)ⁿ.

Solution

P = aⁿr^(0 + 1 + … + (n−1)) = aⁿr^(n(n−1)/2). b = arⁿ⁻¹ ⇒ (ab)ⁿ = a²ⁿrⁿ⁽ⁿ⁻¹⁾ = P².

P²

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

Solution

Sₙ = a(rⁿ − 1)/(r − 1); the next n terms start at arⁿ: arⁿ(rⁿ − 1)/(r − 1). Ratio = 1/rⁿ.

1/rⁿ

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

Solution

b = ar, c = ar², d = ar³. LHS = a²(1 + r² + r⁴) × a²r²(1 + r² + r⁴) = a⁴r²(1 + r² + r⁴)². ab + bc + cd = a²r(1 + r² + r⁴), whose square is the same. Proved.

Proved.

Insert two numbers between 3 and 81 so the sequence is a G.P.

Solution

81 = 3r³ ⇒ r = 3: 9 and 27.

9 and 27

Find n so that (aⁿ⁺¹ + bⁿ⁺¹)/(aⁿ + bⁿ) is the G.M. of a and b.

Solution

aⁿ⁺¹ + bⁿ⁺¹ = √(ab)(aⁿ + bⁿ) ⇒ a^(n+½)(√a − √b) = b^(n+½)(√a − √b) ⇒ (a/b)^(n+½) = 1 = (a/b)⁰ ⇒ n = −1/2.

n = −1/2

The sum of two numbers is 6 times their G.M. Show they are in the ratio (3 + 2√2) : (3 − 2√2).

Solution

a + b = 6√(ab) ⇒ (a + b)/(2√(ab)) = 3. Componendo–dividendo: (√a + √b)²/(√a − √b)² = (3 + 1)/(3 − 1) = 2 ⇒ (√a + √b)/(√a − √b) = √2 ⇒ √a/√b = (√2 + 1)/(√2 − 1). Squaring: a/b = (3 + 2√2)/(3 − 2√2).

See the working above.

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

Solution

a + b = 2A, ab = G²: a, b are roots of x² − 2Ax + G² = 0: x = A ± √(A² − G²) = A ± √((A + G)(A − G)).

A ± √((A + G)(A − G))

Bacteria double every hour; 30 at the start. How many at the end of the 2nd, 4th and nth hour?

Solution

30 × 2ⁿ: 120, 480, 30 × 2ⁿ.

120, 480, 30 × 2ⁿ

What will ₹500 amount to in 10 years at 10% compounded annually?

Solution

500(1.1)¹⁰ ≈ ₹1296.87.

₹1296.87

The A.M. and G.M. of the roots of a quadratic are 8 and 5. Find the equation.

Solution

Sum of roots 16, product 25: x² − 16x + 25 = 0.

x² − 16x + 25 = 0

Miscellaneous Exercise on Chapter 8

f(x + y) = f(x)f(y) for x, y ∈ N, f(1) = 3 and Σ_(x=1)ⁿ f(x) = 120. Find n.

Solution

f(2) = 9, f(3) = 27, … f(x) = 3ˣ. 3(3ⁿ − 1)/2 = 120 ⇒ 3ⁿ = 81 ⇒ n = 4.

n = 4

The sum of some terms of a G.P. is 315; a = 5, r = 2. Find the last term and the number of terms.

Solution

5(2ⁿ − 1) = 315 ⇒ 2ⁿ = 64 ⇒ n = 6; last term 5 × 2⁵ = 160.

n = 6; 160

The first term is 1 and the sum of the 3rd and 5th terms is 90. Find r.

Solution

r² + r⁴ = 90 ⇒ r² = 9 (−10 rejected) ⇒ r = ±3.

r = ±3

Three numbers in G.P. add to 56. Subtracting 1, 7, 21 gives an A.P. Find them.

Solution

a(1 + r + r²) = 56; 2(ar − 7) = (a − 1) + (ar² − 21) ⇒ a(r − 1)² = 8. Dividing: (1 + r + r²)/(r − 1)² = 7 ⇒ 6r² − 15r + 6 = 0 ⇒ r = 2 or ½; a = 8 or 32: 8, 16, 32 (or 32, 16, 8).

8, 16, 32

A G.P. has an even number of terms; the sum of all terms is 5 times the sum of the odd-place terms. Find r.

Solution

2n terms: S = a(r²ⁿ − 1)/(r − 1); odd places a, ar², …: a(r²ⁿ − 1)/(r² − 1). Ratio = r + 1 = 5 ⇒ r = 4.

r = 4

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

Solution

Componendo–dividendo on the first equality: 2a/(2bx) = 2b/(2cx) ⇒ a/b = b/c ⇒ b² = ac; similarly c² = bd. So a/b = b/c = c/d: in G.P.

in G.P.

S = sum, P = product, R = sum of reciprocals of n terms of a G.P. Prove P²Rⁿ = Sⁿ.

Solution

S = a(rⁿ − 1)/(r − 1); P = aⁿr^(n(n−1)/2); R = (1/a)(1 + 1/r + … ) = (rⁿ − 1)/(arⁿ⁻¹(r − 1)). P²Rⁿ = a²ⁿrⁿ⁽ⁿ⁻¹⁾ × (rⁿ − 1)ⁿ/(aⁿrⁿ⁽ⁿ⁻¹⁾(r − 1)ⁿ) = aⁿ(rⁿ − 1)ⁿ/(r − 1)ⁿ = Sⁿ.

Sⁿ

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

Solution

b = ak, c = ak², d = ak³: the three sums are aⁿ(1 + kⁿ), aⁿkⁿ(1 + kⁿ), aⁿk²ⁿ(1 + kⁿ): each is kⁿ times the previous: in G.P.

in G.P.

a, b are roots of x² − 3x + p = 0 and c, d of x² − 12x + q = 0, with a, b, c, d in G.P. Prove (q + p) : (q − p) = 17 : 15.

Solution

a + b = a(1 + r) = 3, c + d = ar²(1 + r) = 12 ⇒ r² = 4. r = 2: a = 1: 1, 2, 4, 8 ⇒ p = ab = 2, q = cd = 32; (q + p) : (q − p) = 34 : 30 = 17 : 15. (r = −2 gives a = −3, p = −18, q = −288 and the same ratio.)

17 : 15

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

Solution

(a + b)/(2√(ab)) = m/n. Componendo–dividendo: (√a + √b)²/(√a − √b)² = (m + n)/(m − n) ⇒ √a/√b = (√(m + n) + √(m − n))/(√(m + n) − √(m − n)). Squaring: a/b = [2m + 2√(m² − n²)]/[2m − 2√(m² − n²)]: proved.

proved

Sum to n terms: (i) 5 + 55 + 555 + … (ii) 0.6 + 0.66 + 0.666 + ….

Solution

(i) (5/9)[9 + 99 + …] = (5/9)[10(10ⁿ − 1)/9 − n] = (50/81)(10ⁿ − 1) − 5n/9. (ii) (6/9)[0.9 + 0.99 + …] = (2/3)[n − (0.1 + 0.01 + …)] = (2/3)[n − (1 − 10⁻ⁿ)/9] = 2n/3 − (2/27)(1 − 10⁻ⁿ).

(50/81)(10ⁿ − 1) − 5n/9; 2n/3 − (2/27)(1 − 10⁻ⁿ)

Find the 20th term of 2 × 4 + 4 × 6 + 6 × 8 + … .

Solution

aₙ = 2n(2n + 2) = 4n(n + 1): a₂₀ = 4 × 20 × 21 = 1680.

1680

A tractor: ₹12000; ₹6000 cash, the rest in annual instalments of ₹500 plus 12% interest on the unpaid amount. Total cost?

Solution

₹6000 unpaid ⇒ 12 instalments; interest on 6000, 5500, …, 500 at 12%: 0.12 × 500(12 + 11 + … + 1) = 0.12 × 500 × 78 = 4680. Total = 12000 + 4680 = ₹16680.

₹16680

A scooter: ₹22000; ₹4000 cash, the balance in annual instalments of ₹1000 plus 10% interest on the unpaid amount. Total cost?

Solution

₹18000 in 18 instalments; interest 0.10 × 1000(18 + 17 + … + 1) = 100 × 171 = 17100. Total = 22000 + 17100 = ₹39100.

₹39100

A chain letter: each person mails 4 copies, 50 paise each. Find the postage when the 8th set is mailed.

Solution

Letters 4 + 4² + … + 4⁸ = 4(4⁸ − 1)/3 = 87380; cost 87380 × ₹0.50 = ₹43690.

₹43690

₹10000 at 5% simple interest. Find the amount in the 15th year and after 20 years.

Solution

Interest ₹500 a year. In the 15th year (after 14 years' interest): 10000 + 14 × 500 = ₹17000; after 20 years: 10000 + 20 × 500 = ₹20000.

₹17000; ₹20000

A machine costs ₹15625 and depreciates 20% each year. Value after 5 years?

Solution

15625 × (0.8)⁵ = ₹5120.

₹5120

150 workers start a job; 4 drop out from the second day, 4 more the third day, and so on; it took 8 more days than planned. Find the days taken.

Solution

Planned n days: work 150n. Actual n + 8 days with 150, 146, …: (n + 8)/2 × [2(150) − 4(n + 7)] = 150n ⇒ (n + 8)(272 − 4n) = 300n ⇒ n² + 15n − 544 = 0 ⇒ n = 17. The work took 25 days.

25 days

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