Sequences and Series Solved Examples (Class 11 Mathematics)
Sequences and series numericals cover arithmetic progressions, geometric progressions, and sum formulas. These problems develop algebraic reasoning essenti
TL;DR: Sequences and series numericals cover arithmetic progressions, geometric progressions, and sum formulas. These problems develop algebraic reasoning es…
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Sequences and series numericals cover arithmetic progressions, geometric progressions, and sum formulas. These problems develop algebraic reasoning essenti
Sequences and Series — Solved Numerical Examples (Step by Step)
Example 1: Find the 15th term of an AP with first term 5 and common difference 3.
Solution: First term a = 5, Common difference d = 3, n = 15. Using an = a + (n-1)d: a₁₅ = 5 + (15-1) × 3 = 5 + 14 × 3 = 5 + 42 = 47.
Example 2: Find the sum of first 20 terms of an AP with first term 2 and last term 59.
Solution: First term a = 2, Last term l = 59, n = 20. Using Sn = n/2 × (a + l): S₂₀ = 20/2 × (2 + 59) = 10 × 61 = 610.
Example 3: Find the 6th term of a GP with first term 2 and common ratio 3.
Solution: First term a = 2, Common ratio r = 3, n = 6. Using an = a × r^(n-1): a₆ = 2 × 3^5 = 2 × 243 = 486.
Example 4: Find the sum of the series 1 + 2 + 4 + 8 + ... up to 10 terms.
Solution: First term a = 1, Common ratio r = 2, n = 10. Using Sn = a × (r^n - 1) / (r - 1): S₁₀ = 1 × (2^10 - 1) / (2 - 1) = (1024 - 1) / 1 = 1023.
Example 5: Find the sum of infinite GP with first term 1 and common ratio 0.5.
Solution: First term a = 1, Common ratio r = 0.5 (|r| < 1, so infinite sum exists). Sum = a / (1 - r) = 1 / (1 - 0.5) = 1 / 0.5 = 2.
Example 6: How many terms of the AP 3, 7, 11, ... sum to 406?
Solution: First term a = 3, Common difference d = 4, Sum Sn = 406. Using Sn = n/2 × (2a + (n-1)d): 406 = n/2 × (6 + 4(n-1)). 812 = n × (6 + 4n - 4) = n × (2 + 4n). 812 = 2n + 4n². 4n² + 2n - 812 = 0. 2n² + n - 406 = 0. Using quadratic formula: n = (-1 ± sqrt(1 + 3248)) / 4 = (-1 ± 57) / 4. n = 14 (taking positive value).
Tips
- AP: an = a + (n-1)d and Sn = n/2 × (2a + (n-1)d).
- GP: an = a × r^(n-1) and Sn = a × (r^n - 1)/(r - 1) for r ≠ 1.
- Infinite GP sum = a/(1-r) exists only when |r| < 1.
Frequently Asked Questions
What is the difference between a sequence and a series?
A sequence is an ordered list of numbers (e.g., 2, 4, 6, 8). A series is the sum of terms in a sequence (e.g., 2 + 4 + 6 + 8 = 20).
Why is the common ratio important in a GP?
The common ratio determines the growth rate. If |r| < 1, the GP converges (infinite sum exists). If |r| > 1, terms grow infinitely. If r = 1, all terms are equal.
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