NCERT Solutions for Class 11 Maths Chapter 8: Sequences and Series – Free PDF Download

Chapter 8, Sequences and Series, builds on the arithmetic progression (A.P.) concepts introduced in Class 10 and takes them further into geometric progressions (G.P.), the relationship between arithmetic mean (A.M.) and geometric mean (G.M.), and sigma notation for series. In the current rationalised NCERT textbook the chapter contains two exercises — Exercise 8.1 on sequences, and Exercise 8.2 covering geometric progressions and the A.M.–G.M. relationship — followed by a Miscellaneous Exercise that combines ideas from both. These Class 11 Maths Chapter 8 solutions are also useful as quick revision notes before exams.

Exercise 8.1

Q1. Write the first five terms of the sequence whose nth term is a_n = n(n + 2).

Substituting 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
The first five terms are 3, 8, 15, 24, 35.

Q2. Write the first five terms of the sequence whose nth term is a_n = n/(n + 1).

a₁ = 1/2, a₂ = 2/3, a₃ = 3/4, a₄ = 4/5, a₅ = 5/6
The first five terms are 1/2, 2/3, 3/4, 4/5, 5/6.

Q3. Write the first five terms of the sequence whose nth term is a_n = 2ⁿ.

a₁ = 2, a₂ = 4, a₃ = 8, a₄ = 16, a₅ = 32
The first five terms are 2, 4, 8, 16, 32.

Q4. Write the first five terms of the sequence whose nth term is a_n = (2n − 3)/6.

a₁ = −1/6, a₂ = 1/6, a₃ = 1/2, a₄ = 5/6, a₅ = 7/6
The first five terms are −1/6, 1/6, 1/2, 5/6, 7/6.

Q5. Write the first five terms of the sequence whose nth term is a_n = (−1)^{n − 1}5^{n + 1}.

a₁ = (−1)⁰5² = 25, a₂ = (−1)¹5³ = −125, a₃ = 5⁴ = 625, a₄ = −5⁵ = −3125, a₅ = 5⁶ = 15625
The first five terms are 25, −125, 625, −3125, 15625.

Q6. Write the first five terms of the sequence whose nth term is a_n = n(n² + 5)/4.

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

Q7. Find the indicated terms in the sequence whose nth term is a_n = 4n − 3; a₁₇, a₂₄.

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

Q8. Find the indicated term in the sequence whose nth term is a_n = n²/2ⁿ; a₇.

a₇ = 7²/2⁷ = 49/128

Q9. Find the indicated term in the sequence whose nth term is a_n = (−1)^{n − 1}n³; a₉.

a₉ = (−1)⁸(9)³ = 1 × 729 = 729

Q10. Find the indicated term in the sequence whose nth term is a_n = n(n − 2)/(n + 3); a₂₀.

a₂₀ = 20(18)/23 = 360/23

Q11. Write the first five terms of the sequence and obtain the corresponding series: a₁ = 3, a_n = 3a_{n − 1} + 2 for all n > 1.

a₁ = 3
a₂ = 3(3) + 2 = 11
a₃ = 3(11) + 2 = 35
a₄ = 3(35) + 2 = 107
a₅ = 3(107) + 2 = 323
The first five terms are 3, 11, 35, 107, 323, and the corresponding series is 3 + 11 + 35 + 107 + 323 + …

Q12. Write the first five terms of the sequence and obtain the corresponding series: a₁ = −1, a_n = a_{n − 1}/n, n ≥ 2.

a₁ = −1
a₂ = −1/2
a₃ = (−1/2)/3 = −1/6
a₄ = (−1/6)/4 = −1/24
a₅ = (−1/24)/5 = −1/120
The first five terms are −1, −1/2, −1/6, −1/24, −1/120, and the series is −1 − 1/2 − 1/6 − 1/24 − 1/120 + …

Q13. Write the first five terms of the sequence and obtain the corresponding series: a₁ = a₂ = 2, a_n = a_{n − 1} − 1, n > 2.

a₁ = 2, a₂ = 2
a₃ = a₂ − 1 = 1
a₄ = a₃ − 1 = 0
a₅ = a₄ − 1 = −1
The first five terms are 2, 2, 1, 0, −1, and the series is 2 + 2 + 1 + 0 + (−1) + …

Q14. The Fibonacci sequence is defined by 1 = a₁ = a₂ and a_n = a_{n − 1} + a_{n − 2}, n > 2. Find a_{n + 1}/a_n, for n = 1, 2, 3, 4, 5.

The Fibonacci terms are a₁ = 1, a₂ = 1, a₃ = 2, a₄ = 3, a₅ = 5, a₆ = 8.

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

Exercise 8.2

Q1. Find the 20th and nth terms of the G.P. 5/2, 5/4, 5/8, …

Here a = 5/2 and common ratio r = (5/4)/(5/2) = 1/2.
a_n = ar^{n − 1} = (5/2)(1/2)^{n − 1} = 5/2ⁿ
So a₂₀ = 5/2²⁰ = 5/1048576, and the nth term is 5/2ⁿ.

Q2. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.

a₈ = ar⁷ = 192, and r = 2, so a(128) = 192 ⟹ a = 3/2.
a₁₂ = ar¹¹ = (3/2)(2048) = 3072

Q3. The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q² = ps.

Let the first term be A and common ratio R.
p = AR⁴, q = AR⁷, s = AR¹⁰
q² = A²R¹⁴
ps = AR⁴ × AR¹⁰ = A²R¹⁴
Since q² = ps, this is proved.

Q4. The 4th term of a G.P. is square of its second term, and the first term is −3. Determine its 7th term.

Let common ratio be r, a = −3.
a₄ = a₂² ⟹ ar³ = (ar)² = a²r² ⟹ r = a (dividing both sides by ar², valid since a, r ≠ 0)
So r = −3.
a₇ = ar⁶ = (−3)(−3)⁶ = (−3)(729) = −2187

Q5. Which term of the following sequences:

1. 2, 2√2, 4, … is 128?
2. √3, 3, 3√3, … is 729?
3. 1/3, 1/9, 1/27, … is 1/19683?

(a) a = 2, r = √2. a_n = 2(√2)^{n − 1} = 128 = 2⁷. Since √2 = 2^1/2, we need 1 + (n − 1)/2 = 7 ⟹ (n − 1)/2 = 6 ⟹ n = 13.
(b) a = √3, r = √3. a_n = (√3)ⁿ = 729 = 3⁶, i.e., 3^n/2 = 3⁶ ⟹ n = 12.
(c) a = 1/3, r = 1/3. a_n = (1/3)ⁿ = 1/19683 = 1/3⁹ ⟹ n = 9.

Q6. For what values of x, the numbers −2/7, x, −7/2 are in G.P.?

Since the numbers are in G.P., x² = (−2/7)(−7/2) = 1, so x = ±1.

Q7. Find the sum to indicated number of terms in the G.P. 0.15, 0.015, 0.0015, … 20 terms.

a = 0.15, r = 0.1
S₂₀ = a(1 − r²⁰)/(1 − r) = 0.15(1 − (0.1)²⁰)/0.9 = (1/6)[1 − (0.1)²⁰]

Q8. Find the sum to indicated number of terms in the G.P. √7, √21, 3√7, … n terms.

a = √7, r = √21/√7 = √3
S_n = a(rⁿ − 1)/(r − 1) = √7[(√3)ⁿ − 1]/(√3 − 1)

Q9. Find the sum to indicated number of terms in the G.P. 1, −a, a², −a³, … n terms (if a ≠ −1).

a (first term) = 1, common ratio = −a
S_n = [1 − (−a)ⁿ]/[1 − (−a)] = [1 − (−a)ⁿ]/(1 + a)

Q10. Find the sum to indicated number of terms in the G.P. x³, x⁵, x⁷, … n terms (if x ≠ ±1).

a = x³, r = x²
S_n = x³(x²ⁿ − 1)/(x² − 1)

Q11. Evaluate Σ_{k = 1}¹¹ (2 + 3^k).

Σ(2 + 3^k) = Σ2 + Σ3^k = 2(11) + [3(3¹¹ − 1)/(3 − 1)]
= 22 + 3(177147 − 1)/2 = 22 + 3(177146)/2 = 22 + 265719 = 265741

Q12. 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.

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

Q13. How many terms of G.P. 3, 3², 3³, … are needed to give the sum 120?

S_n = 3(3ⁿ − 1)/(3 − 1) = 120 ⟹ 3ⁿ − 1 = 80 ⟹ 3ⁿ = 81 = 3⁴ ⟹ n = 4

Q14. 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.

Let a be first term, r common ratio.
a + ar + ar² = a(1 + r + r²) = 16 …(1)
ar³ + ar⁴ + ar⁵ = ar³(1 + r + r²) = 128 …(2)
Dividing (2) by (1): r³ = 8 ⟹ r = 2
Substituting in (1): a(1 + 2 + 4) = 16 ⟹ 7a = 16 ⟹ a = 16/7
S_n = a(rⁿ − 1)/(r − 1) = (16/7)(2ⁿ − 1)

Q15. Given a G.P. with a = 729 and 7th term 64, determine S₇.

a₇ = ar⁶ = 64 ⟹ 729r⁶ = 64 ⟹ r⁶ = 64/729 = (2/3)⁶ ⟹ 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

Q16. Find a G.P. for which sum of the first two terms is −4 and the fifth term is 4 times the third term.

a₅ = 4a₃ ⟹ ar⁴ = 4ar² ⟹ r² = 4 ⟹ r = ±2
Sum: a + ar = a(1 + r) = −4
If r = 2: a(3) = −4 ⟹ a = −4/3. G.P.: −4/3, −8/3, −16/3, …
If r = −2: a(−1) = −4 ⟹ a = 4. G.P.: 4, −8, 16, −32, …

Q17. 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.

Let first term = a, common ratio = r.
x = ar³, y = ar⁹, z = ar¹⁵
y/x = r⁶ and z/y = r⁶
Since y/x = z/y, x, y, z are in G.P.

Q18. Find the sum to n terms of the sequence, 8, 88, 888, 8888… .

S_n = 8 + 88 + 888 + … to n terms = (8/9)(9 + 99 + 999 + … to n terms)
= (8/9)Σ(10^k − 1), k = 1 to n = (8/9)[(10(10ⁿ − 1)/9) − n]
= (8/81)[10^{n + 1} − 10 − 9n]

Q19. 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.

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

Q20. Show that the products of the corresponding terms of the sequences a, ar, ar², … ar^{n − 1} and A, AR, AR², … AR^{n − 1} form a G.P, and find the common ratio.

The products are: aA, ar·AR, ar²·AR², …, ar^{n − 1}AR^{n − 1}, i.e., aA, aA(rR), aA(rR)², …, aA(rR)^{n − 1}.
Since each term is obtained by multiplying the previous term by the constant rR, this sequence is a G.P. with common ratio rR.

Q21. 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.

Let the terms be a, ar, ar², ar³.
ar² − a = 9 ⟹ a(r² − 1) = 9 …(1)
ar − ar³ = 18 ⟹ −ar(r² − 1) = 18 …(2)
Dividing (2) by (1): −r = 2 ⟹ r = −2
From (1): a(4 − 1) = 9 ⟹ a = 3
The four numbers are 3, −6, 12, −24.

Q22. If the p^th, q^th and r^th terms of a G.P. are a, b and c, respectively. Prove that a^{q − r} b^{r − p} c^{p − q} = 1.

Let A be the first term and R the common ratio.
a = AR^{p − 1}, b = AR^{q − 1}, c = AR^{r − 1}
a^{q − r}b^{r − p}c^{p − q} = A^{(q − r) + (r − p) + (p − q)} R^{(p − 1)(q − r) + (q − 1)(r − p) + (r − 1)(p − q)}
The exponent of A is 0. Expanding the exponent of R:
(p − 1)(q − r) + (q − 1)(r − p) + (r − 1)(p − q) = 0 (all terms cancel on expansion)
So the expression equals A⁰R⁰ = 1.

Q23. 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.

P = a · ar · ar² · … · ar^{n − 1} = aⁿr^{0 + 1 + 2 + … + (n − 1)} = aⁿr^{n(n − 1)/2}
P² = a²ⁿr^{n(n − 1)} = [a² r^{n − 1}]ⁿ = [a · ar^{n − 1}]ⁿ = (ab)ⁿ, since b = ar^{n − 1}.

Q24. 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.

S_n (first n terms) = a(rⁿ − 1)/(r − 1)
The terms from (n + 1) to 2n form a G.P. with first term arⁿ and n terms, so their sum = arⁿ(rⁿ − 1)/(r − 1)
Ratio = [a(rⁿ − 1)/(r − 1)] / [arⁿ(rⁿ − 1)/(r − 1)] = 1/rⁿ

Q25. If a, b, c and d are in G.P. show that (a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)².

Let common ratio = r, so b = ar, c = ar², d = ar³.
LHS = a²(1 + r² + r⁴) × a²r²(1 + r² + r⁴) = a⁴r²(1 + r² + r⁴)²
RHS = (a²r + a²r³ + a²r⁵)² = [a²r(1 + r² + r⁴)]² = a⁴r²(1 + r² + r⁴)²
Since LHS = RHS, this is proved.

Q26. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Let the numbers be G₁, G₂ so that 3, G₁, G₂, 81 is a G.P.
81 = 3r³ ⟹ r³ = 27 ⟹ r = 3
G₁ = 3(3) = 9, G₂ = 9(3) = 27

Q27. Find the value of n so that (a^{n + 1} + b^{n + 1})/(aⁿ + bⁿ) may be the geometric mean between a and b.

We need (a^{n + 1} + b^{n + 1})/(aⁿ + bⁿ) = √(ab).
Trying n = −1/2: (a^1/2 + b^1/2)/(a^−1/2 + b^−1/2) = (√a + √b)/[(√a + √b)/√(ab)] = √(ab).
This satisfies the required condition, so n = −1/2.

Q28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3 + 2√2) : (3 − 2√2).

Let the numbers be a and b, with a + b = 6√(ab).
Let a/b = k. Dividing by b: k + 1 = 6√k. Let u = √k: u² − 6u + 1 = 0 ⟹ u = 3 ± 2√2
So k = u² = (3 ± 2√2)². Since (3 + 2√2)(3 − 2√2) = 9 − 8 = 1, we have 3 − 2√2 = 1/(3 + 2√2), so
(3 + 2√2)² = (3 + 2√2)/(3 − 2√2)
Hence a : b = (3 + 2√2) : (3 − 2√2).

Q29. 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)}.

Let the numbers be a and b. Then a + b = 2A and ab = G².
a and b are roots of t² − 2At + G² = 0.
t = [2A ± √(4A² − 4G²)]/2 = A ± √(A² − G²) = A ± √{(A + G)(A − G)}

Q30. 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 the nth hour?

Bacteria count after n hours = 30 × 2ⁿ
End of 2nd hour: 30 × 2² = 120
End of 4th hour: 30 × 2⁴ = 480
End of nth hour: 30 × 2ⁿ

Q31. What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

Amount = 500(1 + 0.1)¹⁰ = 500(1.1)¹⁰ ≈ Rs 1296.87

Q32. If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.

Sum of roots = 2 × A.M. = 2(8) = 16
Product of roots = (G.M.)² = 5² = 25
The quadratic equation is x² − 16x + 25 = 0

Miscellaneous Exercise

Q1. If f is a function satisfying f(x + y) = f(x) f(y) for all x, y ∈ N such that f(1) = 3 and Σ_{x = 1}ⁿ f(x) = 120, find the value of n.

Since f(1) = 3 and f(x + y) = f(x)f(y), it follows that f(n) = 3ⁿ for all natural numbers n (f is geometric).
Σ_{x = 1}ⁿ 3^x = 3(3ⁿ − 1)/(3 − 1) = 120 ⟹ 3ⁿ − 1 = 80 ⟹ 3ⁿ = 81 = 3⁴ ⟹ n = 4

Q2. The sum of some terms of G.P. is 315 whose first term and common ratio are 5 and 2, respectively. Find the last term and the number of terms.

S_n = 5(2ⁿ − 1)/(2 − 1) = 315 ⟹ 2ⁿ − 1 = 63 ⟹ 2ⁿ = 64 = 2⁶ ⟹ n = 6
Last term = ar^{n − 1} = 5 × 2⁵ = 160

Q3. 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.

a = 1. ar² + ar⁴ = 90 ⟹ r² + r⁴ = 90
Let y = r²: y² + y − 90 = 0 ⟹ y = (−1 ± 19)/2 ⟹ y = 9 or y = −10 (rejected, since y = r² ≥ 0)
r² = 9 ⟹ r = ±3

Q4. 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.

Let the numbers be a/r, a, ar.
Sum: a/r + a + ar = 56 …(1)
After subtracting: a/r − 1, a − 7, ar − 21 are in A.P., so 2(a − 7) = (a/r − 1) + (ar − 21)
2a − 14 = a/r + ar − 22 ⟹ 2a + 8 = a(1/r + r) …(2)
From (1): a(1/r + r) = 56 − a. Substituting into (2): 2a + 8 = 56 − a ⟹ 3a = 48 ⟹ a = 16
Then 16(1/r + r) = 40 ⟹ 1/r + r = 5/2 ⟹ 2r² − 5r + 2 = 0 ⟹ r = 2 or r = 1/2
The numbers are 8, 16, 32 (or the same set in reverse order for r = 1/2).

Q5. 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.

Let the G.P. have 2n terms: a, ar, ar², …, ar^{2n − 1}
Sum of all terms = a(r²ⁿ − 1)/(r − 1)
Sum of terms at odd places (a, ar², ar⁴, …, ar^{2n − 2}) = a(r²ⁿ − 1)/(r² − 1)
Given: a(r²ⁿ − 1)/(r − 1) = 5 × a(r²ⁿ − 1)/(r² − 1)
1/(r − 1) = 5/[(r − 1)(r + 1)] ⟹ r + 1 = 5 ⟹ r = 4

Q6. 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.

From the first equality: (a + bx)(b − cx) = (b + cx)(a − bx)
Expanding and simplifying gives 2x(b² − ac) = 0. Since x ≠ 0, b² = ac.
Similarly, from the second equality, c² = bd.
From b² = ac, b/a = c/b, and from c² = bd, c/b = d/c. Since c/b is common to both, b/a = c/b = d/c, so a, b, c, d are in G.P.

Q7. 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.

For a G.P. a, ar, …, ar^{n − 1}:
S = a(rⁿ − 1)/(r − 1)
P = aⁿr^{n(n − 1)/2}
R = (rⁿ − 1)/[ar^{n − 1}(r − 1)] (sum of reciprocals, which is itself a G.P. with first term 1/a and ratio 1/r)
P²Rⁿ = a²ⁿr^{n(n − 1)} × (rⁿ − 1)ⁿ/[aⁿr^{n(n − 1)}(r − 1)ⁿ] = aⁿ(rⁿ − 1)ⁿ/(r − 1)ⁿ = [a(rⁿ − 1)/(r − 1)]ⁿ = Sⁿ

Q8. If a, b, c, d are in G.P, prove that (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in G.P.

Let b = ar, c = ar², d = ar³.
aⁿ + bⁿ = aⁿ(1 + rⁿ)
bⁿ + cⁿ = aⁿrⁿ(1 + rⁿ)
cⁿ + dⁿ = aⁿr²ⁿ(1 + rⁿ)
These three terms have common ratio rⁿ, so they are in G.P.

Q9. 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.

a + b = 3, ab = p; c + d = 12, cd = q
Let a, b, c, d be A, Ar, Ar², Ar³ (consecutive terms of the G.P.).
a + b = A(1 + r) = 3, c + d = Ar²(1 + r) = 12
Dividing: r² = 4
p = ab = A²r, q = cd = A²r⁵
(q + p)/(q − p) = (r⁴ + 1)/(r⁴ − 1) = (16 + 1)/(16 − 1) = 17/15
Hence (q + p) : (q − p) = 17 : 15.

Q10. 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)).

(a + b)/(2√(ab)) = m/n. Let t = √(a/b). Then n(t² + 1) = 2mt ⟹ nt² − 2mt + n = 0
t = [m ± √(m² − n²)]/n
Let t₁ = [m + √(m² − n²)]/n, t₂ = [m − √(m² − n²)]/n. Since t₁t₂ = (m² − (m² − n²))/n² = 1, t₂ = 1/t₁.
a/b = t₁² = t₁/t₂ = [m + √(m² − n²)]/[m − √(m² − n²)]
Hence a : b = (m + √(m² − n²)) : (m − √(m² − n²)).

Q11. Find the sum of the following series up to n terms:

1. 5 + 55 + 555 + …
2. 0.6 + 0.66 + 0.666 + …

(i) S_n = 5 + 55 + 555 + … = (5/9)(9 + 99 + 999 + …) = (5/9)Σ(10^k − 1), k = 1 to n
= (5/9)[(10^{n + 1} − 10)/9 − n] = (5/81)[10^{n + 1} − 10 − 9n]

(ii) S_n = 0.6 + 0.66 + 0.666 + … = (6/9)[(1 − 0.1) + (1 − 0.01) + … + (1 − 10⁻ⁿ)]
= (2/3)[n − (1 − 10⁻ⁿ)/9] = (2/3)n − (2/27)(1 − 10⁻ⁿ)

Q12. Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + … + n terms.

The nth term is a_n = (2n)(2n + 2) = 4n(n + 1)
a₂₀ = 4(20)(21) = 1680

Q13. 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?

Balance = Rs 6000, paid in 12 instalments of Rs 500 each.
Interest is paid each year on the unpaid amount: 12% of 6000 = 720, then 12% of 5500 = 660, and so on, decreasing by 60 each year, down to 12% of 500 = 60.
Total interest = sum of A.P. with a = 720, d = −60, n = 12, last term = 60
Sum = (12/2)(720 + 60) = 6 × 780 = 4680
Total cost = Rs 6000 (cash) + Rs 6000 (instalment principal) + Rs 4680 (interest) = Rs 16680

Q14. 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?

Balance = Rs 18000, paid in 18 instalments of Rs 1000 each.
Interest: 10% of 18000 = 1800, then 10% of 17000 = 1700, decreasing by 100 each year down to 10% of 1000 = 100.
Total interest = sum of A.P. with a = 1800, d = −100, n = 18, last term = 100
Sum = (18/2)(1800 + 100) = 9 × 1900 = 17100
Total cost = Rs 4000 + Rs 18000 + Rs 17100 = Rs 39100

Q15. 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 letters is mailed.

Number of letters in the kth set = 4^k (1st set: 4, 2nd set: 16, and so on)
Total letters after 8 sets = Σ_{k = 1}⁸ 4^k = 4(4⁸ − 1)/(4 − 1) = 4(65536 − 1)/3 = 87380
Amount spent = 87380 × Rs 0.50 = Rs 43690

Q16. 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.

Simple interest per year = 5% of 10000 = Rs 500
Amount in 15th year (principal plus interest for 14 completed years) = 10000 + 14 × 500 = Rs 17000
Total amount after 20 years = 10000 + 20 × 500 = Rs 20000

Q17. 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.

Value after 5 years = 15625 × (1 − 0.2)⁵ = 15625 × (4/5)⁵ = 15625 × 1024/3125 = 5 × 1024 = Rs 5120

Q18. 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.

Let n be the number of days planned originally (with 150 workers each day), so total work = 150n man-days.
With dropouts, workers on day k = 150 − 4(k − 1), for k = 1 to (n + 8) days.
Total work done = Σ_k=1ⁿ⁺⁸[150 − 4(k − 1)] = 150(n + 8) − 2(n + 7)(n + 8)
Setting this equal to 150n: 150(n + 8) − 2(n + 7)(n + 8) = 150n
1200 = 2(n + 7)(n + 8) ⟹ (n + 7)(n + 8) = 600 ⟹ n² + 15n − 544 = 0
n = (−15 + √2401)/2 = (−15 + 49)/2 = 17
The work was actually completed in n + 8 = 25 days.

Class 11 Maths Chapter 8 – Notes and Extra Questions

• nth term of an A.P.: a_n = a + (n − 1)d, where a is the first term and d is the common difference.
• Sum of first n terms of an A.P.: S_n = (n/2)[2a + (n − 1)d].
• nth term of a G.P.: a_n = ar^{n − 1}, where a is the first term and r is the common ratio.
• Sum of first n terms of a G.P.: S_n = a(rⁿ − 1)/(r − 1) when r ≠ 1, and S_n = na when r = 1.
• Sum to infinity of a G.P. (when |r| < 1): S_∞ = a/(1 − r).
• Geometric mean (G.M.) of two positive numbers a and b is √(ab), i.e., a, G, b are in G.P.
• Arithmetic mean (A.M.) of two numbers a and b is (a + b)/2.
• Relationship between A.M. and G.M.: A ≥ G for any two positive real numbers, with equality only when the numbers are equal.
• Sigma notation: the series a₁ + a₂ + a₃ + … + a_n is written compactly as Σ_k=1ⁿ a_k.
• To insert n geometric means between a and b, use common ratio r = (b/a)^{1/(n + 1)}.

Frequently Asked Questions

How many exercises are there in Class 11 Maths Chapter 8, Sequences and Series, in the current NCERT textbook?

The current (2026-27 reprint) rationalised NCERT textbook has two exercises in this chapter — Exercise 8.1 (14 questions, on sequences) and Exercise 8.2 (32 questions, on geometric progressions and the A.M.–G.M. relationship) — followed by a Miscellaneous Exercise (18 questions). Note that arithmetic progression (A.P.) is no longer taught as a separate section in this chapter since it is covered in Class 10, and the earlier section on special series (sums of natural numbers, their squares and cubes) has also been removed in the rationalised edition.

What is the difference between a sequence and a series?

A sequence is an ordered list of numbers following a definite rule, such as a₁, a₂, a₃, …, a_n. A series is the sum of the terms of a sequence, written as a₁ + a₂ + a₃ + … + a_n. A sequence gives individual terms, while a series gives their total.

What is the formula for the sum of an infinite geometric progression?

For a G.P. with first term a and common ratio r, where |r| < 1, the sum to infinity is given by S_∞ = a/(1 − r). This formula only applies when the common ratio’s absolute value is less than 1, otherwise the sum does not converge.

How do you find the geometric mean between two numbers?

The geometric mean (G.M.) between two positive numbers a and b is √(ab). This ensures that a, G, b form a geometric progression, since G/a = b/G = √(b/a).