Sequences and Series — NCERT Class 11 Maths

How to Use This Page

This page provides complete, step-by-step NCERT solutions for every exercise in Chapter 8 — Sequences and Series. Work through each solution on your own first, then compare your approach with the steps shown here. Focus on understanding the formula applied, the substitution made, and the final simplification.

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Key Concepts (Quick Recap)

• Arithmetic Progression (AP): aₙ = a + (n − 1)d
• Sum of AP: Sₙ = n/2 [2a + (n − 1)d] = n/2 [a + l]
• Geometric Progression (GP): aₙ = arⁿ⁻¹
• Sum of GP: Sₙ = a(rⁿ − 1)/(r − 1) for r > 1; Sₙ = a(1 − rⁿ)/(1 − r) for |r| < 1
• Arithmetic Mean (AM): AM = (a + b)/2
• Geometric Mean (GM): GM = √(ab)
• AM-GM Inequality: AM ≥ GM (for positive numbers, equality when a = b)
• Sum of first n natural numbers: n(n + 1)/2
• Sum of squares: n(n + 1)(2n + 1)/6
• Sum of cubes: [n(n + 1)/2]²

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Exercise 8.1

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

Substitute n = 1, 2, 3, 4, 5 in aₙ = n(n + 2):

• a₁ = 1(1 + 2) = 1 × 3 = 3
• a₂ = 2(2 + 2) = 2 × 4 = 8
• a₃ = 3(3 + 2) = 3 × 5 = 15
• a₄ = 4(4 + 2) = 4 × 6 = 24
• a₅ = 5(5 + 2) = 5 × 7 = 35

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

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

Substitute n = 1, 2, 3, 4, 5:

• a₁ = 1/(1 + 1) = 1/2
• a₂ = 2/(2 + 1) = 2/3
• a₃ = 3/(3 + 1) = 3/4
• a₄ = 4/(4 + 1) = 4/5
• a₅ = 5/(5 + 1) = 5/6

The first five terms are 1/2, 2/3, 3/4, 4/5, 5/6.

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Q3. Write the first five terms of the sequence whose nth term is aₙ = (−1)ⁿ⁻¹ × 5ⁿ⁺¹.

• a₁ = (−1)⁰ × 5² = 1 × 25 = 25
• a₂ = (−1)¹ × 5³ = −125 = −125
• a₃ = (−1)² × 5⁴ = 1 × 625 = 625
• a₄ = (−1)³ × 5⁵ = −3125 = −3125
• a₅ = (−1)⁴ × 5⁶ = 1 × 15625 = 15625

The first five terms are 25, −125, 625, −3125, 15625.

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Q4. Find the 10th term of the AP: 5, 11, 17, 23, …

Here a = 5, d = 11 − 5 = 6, n = 10.

Using aₙ = a + (n − 1)d:

a₁₀ = 5 + (10 − 1) × 6 = 5 + 54 = 59

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Q5. Find the 12th term of the AP: √2, 3√2, 5√2, 7√2, …

Here a = √2, d = 3√2 − √2 = 2√2, n = 12.

a₁₂ = √2 + (12 − 1)(2√2) = √2 + 22√2 = 23√2 = 23√2

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Q6. Is 301 a term of the AP 5, 11, 17, 23, …?

Here a = 5, d = 6. Let aₙ = 301.

5 + (n − 1)(6) = 301

6(n − 1) = 296

n − 1 = 296/6 = 49.33…

Since n is not a natural number, 301 is not a term of this AP.

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Exercise 8.2

Q1. Find the sum of the first 20 terms of the AP: 1, 4, 7, 10, …

Here a = 1, d = 3, n = 20.

Using Sₙ = n/2 [2a + (n − 1)d]:

S₂₀ = 20/2 [2(1) + (20 − 1)(3)] = 10 [2 + 57] = 10 × 59 = 590

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Q2. Find the sum of the first 15 terms of the AP: 3, 8, 13, 18, …

Here a = 3, d = 5, n = 15.

S₁₅ = 15/2 [2(3) + (15 − 1)(5)] = 15/2 [6 + 70] = 15/2 × 76 = 15 × 38 = 570

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Q3. Find the sum: 7 + 10¹/₂ + 14 + … + 84.

Here a = 7, d = 10.5 − 7 = 3.5 = 7/2, l = 84.

First find n: a + (n − 1)d = 84

7 + (n − 1)(7/2) = 84

(n − 1)(7/2) = 77

n − 1 = 22, so n = 23.

Sₙ = n/2 (a + l) = 23/2 (7 + 84) = 23/2 × 91 = 23 × 45.5 = 1046.5

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Q4. Find the sum of the first 10 terms of the GP: 3, 6, 12, 24, …

Here a = 3, r = 2, n = 10.

Using Sₙ = a(rⁿ − 1)/(r − 1):

S₁₀ = 3(2¹⁰ − 1)/(2 − 1) = 3(1024 − 1)/1 = 3 × 1023 = 3069

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Q5. Find the sum of the first 7 terms of the GP: 1, −1/2, 1/4, −1/8, …

Here a = 1, r = −1/2, n = 7.

Using Sₙ = a(1 − rⁿ)/(1 − r):

S₇ = 1[1 − (−1/2)⁷]/[1 − (−1/2)] = [1 − (−1/128)]/[3/2] = [1 + 1/128]/(3/2) = (129/128) × (2/3) = 258/384 = 43/64

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Q6. Find the 5th term of the GP: 2, 2√2, 4, 4√2, …

Here a = 2, r = 2√2/2 = √2.

a₅ = ar⁴ = 2(√2)⁴ = 2 × 4 = 8

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Q7. Insert two arithmetic means between 3 and 27.

Let the two arithmetic means be A₁ and A₂. Then 3, A₁, A₂, 27 are in AP.

Total terms = 4, so a = 3, a₄ = 27.

a₄ = a + 3d → 3 + 3d = 27 → d = 8.

A₁ = a + d = 3 + 8 = 11

A₂ = a + 2d = 3 + 16 = 19

The arithmetic means are 11 and 19.

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Q8. Insert three geometric means between 1 and 256.

Let the geometric means be G₁, G₂, G₃. Then 1, G₁, G₂, G₃, 256 are in GP.

Total terms = 5, so a = 1, a₅ = 256.

a₅ = ar⁴ → 1 × r⁴ = 256 → r⁴ = 256 → r = 4.

G₁ = ar = 4, G₂ = ar² = 16, G₃ = ar³ = 64.

The geometric means are 4, 16, 64.

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Q9. If AM and GM of two numbers are 13 and 12 respectively, find the numbers.

AM = (a + b)/2 = 13 → a + b = 26

GM = √(ab) = 12 → ab = 144

The numbers are roots of x² − 26x + 144 = 0.

x = [26 ± √(676 − 576)]/2 = [26 ± √100]/2 = [26 ± 10]/2

x = 18 or x = 8.

The numbers are 18 and 8.

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Q10. Find the sum: Σ(n = 1 to 30) [1/2 − 1/3 + 1/4 − … terms].

This is the sum of the first n terms of the series with aₙ = 1/(n + 1).

S₃₀ = 1/2 + 1/3 + 1/4 + … + 1/31

This equals H₃₁ − 1 where H₃₁ is the 31st harmonic number.

≈ ln(31) + γ − 1 ≈ 3.43 + 0.577 − 1 ≈ 3.01

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Miscellaneous Exercise

Q1. If the sum of a number and its reciprocal is 2, find the number.

Let the number be x. Then x + 1/x = 2.

x² − 2x + 1 = 0 → (x − 1)² = 0 → x = 1

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Q2. Find the sum of n terms of the series: 1 × 2 + 2 × 3 + 3 × 4 + …

General term: Tₖ = k(k + 1) = k² + k

Sₙ = Σk² + Σk = n(n+1)(2n+1)/6 + n(n+1)/2

= n(n+1)/2 [(2n+1)/3 + 1] = n(n+1)/2 × (2n+4)/3 = n(n+1)(n+2)/3

Sₙ = n(n+1)(n+2)/3

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Q3. Find the sum of n terms of: 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

General term: Tₖ = k(k+1)(k+2) = k³ + 3k² + 2k

Sₙ = Σk³ + 3Σk² + 2Σk

= [n(n+1)/2]² + 3[n(n+1)(2n+1)/6] + 2[n(n+1)/2]

= n(n+1)/4 [n(n+1) + 2(2n+1) + 4]

= n(n+1)/4 [n² + n + 4n + 2 + 4] = n(n+1)/4 [n² + 5n + 6]

= n(n+1)(n+2)(n+3)/4

Sₙ = n(n+1)(n+2)(n+3)/4

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Q4. Find the sum to n terms: 3 × 1² + 5 × 2² + 7 × 3² + …

General term: Tₖ = (2k + 1)k² = 2k³ + k²

Sₙ = 2Σk³ + Σk² = 2[n(n+1)/2]² + n(n+1)(2n+1)/6

= n²(n+1)²/2 + n(n+1)(2n+1)/6

= n(n+1)/6 [3n(n+1) + (2n+1)] = n(n+1)/6 [3n² + 3n + 2n + 1]

= n(n+1)(3n² + 5n + 1)/6

Sₙ = n(n+1)(3n² + 5n + 1)/6

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Q5. The ratio of the sums of m and n terms of an AP is m² : n². Find the ratio of the mth and nth terms.

Sₘ/Sₙ = m²/n² → [m/2(2a + (m−1)d)] / [n/2(2a + (n−1)d)] = m²/n²

(2a + (m−1)d) / (2a + (n−1)d) = m/n

For the ratio of mth to nth term:

aₘ/aₙ = [a + (m−1)d] / [a + (n−1)d]

Let a = kd for some constant k. Then:

Sₘ/Sₙ = m(2k + m − 1) / n(2k + n − 1) = m²/n²

This gives: 2k + m − 1 = m(2k + n − 1)/n ... Solving gives k = 1/2 (i.e., a = d/2).

Then aₘ/aₙ = [d/2 + (m−1)d] / [d/2 + (n−1)d] = (2m − 1)/(2n − 1)

The ratio is (2m − 1) : (2n − 1).

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Q6. If the sum of p terms of an AP is q and the sum of q terms is p, find the sum of (p + q) terms.

Sₚ = p/2[2a + (p−1)d] = q, and Sᵩ = q/2[2a + (q−1)d] = p.

From these: 2a + (p−1)d = 2q/p and 2a + (q−1)d = 2p/q.

Subtracting: (p − q)d = 2(q/p − p/q) = 2(q² − p²)/(pq)

d = −2(p + q)/(pq)

Also, 2a = 2q/p − (p−1)d = 2q/p + 2(p−1)(p+q)/(pq) = [2q² + 2(p−1)(p+q)]/(pq)

S(p+q) = (p+q)/2 [2a + (p+q−1)d]

After substitution and simplification: S(p+q) = −(p + q)

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Q7. Find the sum to infinity of the GP: 1, 1/3, 1/9, 1/27, …

S∞ = a/(1 − r) = 1/(1 − 1/3) = 1/(2/3) = 3/2

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Q8. Find the sum to infinity of the GP: 8, 4√2, 4, 2√2, …

S∞ = a/(1 − r) = 8/(1 − 1/√2) = 8√2/(√2 − 1) × (√2 + 1)/(√2 + 1) = 8√2(√2 + 1)/(2 − 1) = 8√2(√2 + 1) = 16 + 8√2

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Q9. If a, b, c are in GP and a^(1/x) = b^(1/y) = c^(1/z), prove that x, y, z are in AP.

Since a, b, c are in GP: b² = ac → b/a = c/b = r (common ratio).

Let a^(1/x) = b^(1/y) = c^(1/z) = k.

Then a = kˣ, b = kʸ, c = kᶻ.

Since b² = ac: k²ʸ = kˣ × kᶻ = kˣ⁺ᶻ.

Therefore 2y = x + z, which means x, y, z are in AP. Proved.

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Q10. Find the sum of n terms of the series: 5 + 11 + 19 + 29 + 41 + …

Differences: 11 − 5 = 6, 19 − 11 = 8, 29 − 19 = 10, 41 − 29 = 12.

Second differences are constant (= 2), so it is a quadratic sequence.

Let Tₙ = an² + bn + c.

T₁ = a + b + c = 5, T₂ = 4a + 2b + c = 11, T₃ = 9a + 3b + c = 19.

From T₂ − T₁: 3a + b = 6. From T₃ − T₂: 5a + b = 8.

Subtracting: 2a = 2, so a = 1, b = 3, c = 1.

Tₙ = n² + 3n + 1

Sₙ = Σn² + 3Σn + n = n(n+1)(2n+1)/6 + 3n(n+1)/2 + n

= n/6 [(n+1)(2n+1) + 9(n+1) + 6] = n/6 [2n² + 3n + 1 + 9n + 9 + 6]

= n/6 [2n² + 12n + 16] = n(n² + 6n + 8)/3

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Q11. The product of three numbers in GP is 216 and their sum is 19. Find the numbers.

Let the numbers be a/r, a, ar.

Product: (a/r)(a)(ar) = a³ = 216 → a = 6.

Sum: 6/r + 6 + 6r = 19 → 6(1/r + 1 + r) = 19 → 1/r + 1 + r = 19/6

Multiplying by r: 1 + r + r² = 19r/6

6r² + 6r + 6 = 19r → 6r² − 13r + 6 = 0

r = [13 ± √(169 − 144)]/12 = [13 ± 5]/12

r = 18/12 = 3/2 or r = 8/12 = 2/3.

For r = 3/2: numbers are 4, 6, 9. For r = 2/3: numbers are 9, 6, 4.

The numbers are 4, 6, 9.

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Q12. If the AM of two positive numbers a and b is 3 more than their GM, find the ratio a : b.

AM − GM = 3 → (a + b)/2 − √(ab) = 3

Let √a = x, √b = y. Then (x² + y²)/2 − xy = 3 → (x − y)²/2 = 3 → (x − y)² = 6.

Also, (x² + y²)/2 + xy = (x + y)²/2.

We have x² + y² = 2xy + 6 and (x² + y²)/2 = xy + 3.

Since a + b = 2(GM + 3) and ab = GM², let GM = g.

Then a + b = 2g + 6 and ab = g².

From the quadratic x² − (2g + 6)x + g² = 0, the discriminant must be ≥ 0.

(2g + 6)² − 4g² ≥ 0 → 4g² + 24g + 36 − 4g² ≥ 0 → 24g + 36 ≥ 0 → g ≥ −3/2.

Since g > 0, this is always satisfied. Let g = 4 (taking a specific case).

Then a + b = 14, ab = 16 → a, b = 2, 8. Ratio = 1 : 4 (general: (√2 + 1)² : (√2 − 1)²).

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Internal Links

• [Chapter 8 Guide — Sequences and Series](/school/cbse/class/11/maths/cbse-sequences-and-series)
• [All Class 11 Maths Solutions](/school/cbse/class/11/maths/solutions)
• [Class 11 Maths Subject Guide](/school/cbse/class/11/maths)