Sequences and Series — Class 11 Mathematics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Sequences and Series" — 8 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Sequences and Series" — 8 important questions with detailed answers for CBSE board…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

Find the 15th term of the arithmetic progression (AP) 2, 5, 8, 11, ...
The sum of first n terms of an AP is Sₙ = n²/2 + 3n/2. Find the AP and its 20…
Find the sum of the first 20 terms of the geometric progression (GP) 3, 6, 12…
If the sum of an AP with 10 terms is 250 and the first term is 4, find the co…
The ratio of the 5th and 8th terms of a GP is 8:27. If the 6th term is 24, fi…
Find the sum of the infinite GP: 1 + 1/2 + 1/4 + 1/8 + ...
+ 2 more questions in the full chapter

Solutions Summary:

Question	Status
Find the 15th term of the arithmetic progression (AP) 2, …	✓ Solved
The sum of first n terms of an AP is Sₙ = n²/2 + 3n/2. Fi…	✓ Solved
Find the sum of the first 20 terms of the geometric progr…	✓ Solved
If the sum of an AP with 10 terms is 250 and the first te…	✓ Solved
The ratio of the 5th and 8th terms of a GP is 8:27. If th…	✓ Solved
Find the sum of the infinite GP: 1 + 1/2 + 1/4 + 1/8 + ...	✓ Solved

Showing 6 of 8 questions

Q1: Find the 15th term of the arithmetic progression (AP) 2, 5, 8, 11, ...

Given AP: 2, 5, 8, 11, ... First term a = 2 Common difference d = 5 - 2 = 3 Step 1: Use the formula for nth term of an AP: aₙ = a + (n - 1)d Step 2: Substitute n = 15, a = 2, d = 3: a₁₅ = 2 + (15 - 1) × 3 a₁₅ = 2 + 14 × 3 a₁₅ = 2 + 42 a₁₅ = 44 Final Answer: The 15th term is 44

Q2: The sum of first n terms of an AP is Sₙ = n²/2 + 3n/2. Find the AP and its 20th term.

Given: Sₙ = n²/2 + 3n/2 Step 1: Find the first term a₁ using S₁: S₁ = (1)²/2 + 3(1)/2 = 1/2 + 3/2 = 2 So a₁ = 2 Step 2: Find a₂ using S₂: S₂ = (2)²/2 + 3(2)/2 = 2 + 3 = 5 So a₁ + a₂ = 5 Therefore a₂ = 5 - 2 = 3 Step 3: Find common difference d: d = a₂ - a₁ = 3 - 2 = 1 Step 4: Verify with a₃: S₃ = (3)²/2 + 3(3)/2 = 9/2 + 9/2 = 9 a₃ = S₃ - S₂ = 9 - 5 = 4 Common difference d = 4 - 3 = 1 ✓ Step 5: The AP is: 2, 3, 4, 5, ... For the 20th term: a₂₀ = 2 + (20 - 1) × 1 = 2 + 19 = 21 Final Answer: ...

Q3: Find the sum of the first 20 terms of the geometric progression (GP) 3, 6, 12, 24, ...

Given GP: 3, 6, 12, 24, ... First term a = 3 Common ratio r = 6/3 = 2 Step 1: Use the formula for sum of n terms of a GP (r ≠ 1): Sₙ = a(rⁿ - 1)/(r - 1) Step 2: Substitute n = 20, a = 3, r = 2: S₂₀ = 3(2²⁰ - 1)/(2 - 1) S₂₀ = 3(2²⁰ - 1)/1 S₂₀ = 3(1048576 - 1) S₂₀ = 3 × 1048575 S₂₀ = 3145725 Final Answer: Sum of first 20 terms = 3145725

Q4: If the sum of an AP with 10 terms is 250 and the first term is 4, find the common difference and the last term.

Given: n = 10, Sₙ = 250, a = 4 Step 1: Use the sum formula: Sₙ = n/2[2a + (n - 1)d] 250 = 10/2[2(4) + (10 - 1)d] 250 = 5[8 + 9d] 50 = 8 + 9d 9d = 42 d = 14/3 Step 2: Find the last term (a₁₀): aₙ = a + (n - 1)d a₁₀ = 4 + (10 - 1) × 14/3 a₁₀ = 4 + 9 × 14/3 a₁₀ = 4 + 42 a₁₀ = 46 Final Answer: Common difference d = 14/3, Last term = 46

Q5: The ratio of the 5th and 8th terms of a GP is 8:27. If the 6th term is 24, find the first term and common ratio.

Let a be the first term and r be the common ratio. Step 1: Write the given terms: a₅ = ar⁴ a₈ = ar⁷ Ratio: a₅/a₈ = ar⁴/ar⁷ = 1/r³ = 8/27 Step 2: Solve for r: 1/r³ = 8/27 r³ = 27/8 r = 3/2 Step 3: Use the condition a₆ = 24: a₆ = ar⁵ = 24 a(3/2)⁵ = 24 a × 243/32 = 24 a = 24 × 32/243 a = 768/243 a = 256/81 Step 4: Verify: a₅ = (256/81) × (3/2)⁴ = (256/81) × (81/16) = 16 a₈ = (256/81) × (3/2)⁷ = (256/81) × (2187/128) = 54 Ratio = 16:54 = 8:27 ✓ Final Answer: First term a = 256/81, Common ratio ...

Q6: Find the sum of the infinite GP: 1 + 1/2 + 1/4 + 1/8 + ...

Given infinite GP: 1 + 1/2 + 1/4 + 1/8 + ... First term a = 1 Common ratio r = 1/2 Step 1: Check if |r| < 1: |1/2| = 0.5 < 1, so the series converges Step 2: Use the formula for sum of infinite GP (|r| < 1): S∞ = a/(1 - r) Step 3: Substitute values: S∞ = 1/(1 - 1/2) S∞ = 1/(1/2) S∞ = 2 Final Answer: Sum of the infinite GP = 2

Showing 6 of 8 questions. Visit the full page for complete solutions.