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Class 11 · Mathematics NCERT Class 11 Mathematics · Ch. 83 min read · 15 questions

Sequences and Series

Mathematics

Sequences and Series

A sequence is an ordered list of numbers following a definite pattern. Each number in the list is called a term. A series is the sum of the terms of a sequence. Understanding sequences and series is essential for many areas of mathematics, including calculus and discrete mathematics.

  • Types of Sequences:
  • Arithmetic Progression (AP): Each term is obtained by adding a fixed number (called the common difference, d) to the previous term.
  • Geometric Progression (GP): Each term is obtained by multiplying the previous term by a fixed number (called the common ratio, r).

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Key Formulas

  • Arithmetic Progression:
  • General term: an = a + (n - 1)d, where a = first term, d = common difference
  • Sum of n terms: Sn = n/2 × [2a + (n - 1)d] or Sn = n/2 × (a + l), where l = last term
  • Geometric Progression:
  • General term: an = a × rn-1, where a = first term, r = common ratio
  • Sum of n terms: Sn = a(rn - 1)/(r - 1) when r ≠ 1; Sn = na when r = 1
  • Sum of infinite GP (|r| < 1): S = a/(1 - r)
  • Special Sums:
  • Sum of first n natural numbers: 1 + 2 + 3 + ... + n = n(n+1)/2
  • Sum of squares: 12 + 22 + ... + n2 = n(n+1)(2n+1)/6
  • Sum of cubes: 13 + 23 + ... + n3 = [n(n+1)/2]2

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Worked Examples

Example 1

Find the 20th term of the AP: 3, 7, 11, 15, ...
Here a = 3, d = 7 - 3 = 4, n = 20.
a20 = 3 + (20 - 1) × 4 = 3 + 76 = 79

Example 2

Find the sum of first 15 terms of the AP: 5, 8, 11, ...
a = 5, d = 3, n = 15.
S15 = 15/2 × [2(5) + 14(3)] = 15/2 × [10 + 42] = 15/2 × 52 = 390

Example 3

The 3rd and 6th terms of a GP are 12 and 96. Find the first term and common ratio.
a3 = ar2 = 12 and a6 = ar5 = 96.
Dividing: r3 = 96/12 = 8, so r = 2. Then a(4) = 12, a = 3. First term = 3, r = 2

Example 4

Find the sum of the infinite GP: 1 + 1/3 + 1/9 + ...
a = 1, r = 1/3. Since |r| < 1, S = 1/(1 - 1/3) = 1/(2/3) = 3/2

Example 5

Find the sum: 12 + 22 + 32 + ... + 102.
Using formula: n(n+1)(2n+1)/6 with n = 10:
= 10 × 11 × 21 / 6 = 2310/6 = 385

Example 6

If a, b, c are in AP and a, b, c are also in GP, find the values.
In AP: b - a = c - b, so 2b = a + c.
In GP: b2 = ac.
From AP: a + c = 2b. From GP: ac = b2.
This means a = b = c (constant sequence) is the simplest solution.

Example 7

Find three numbers in AP whose sum is 15 and product is 80.
Let numbers be a - d, a, a + d. Sum = 3a = 15, so a = 5.
Product = (5 - d)(5)(5 + d) = 5(25 - d2) = 80. So 25 - d2 = 16, d2 = 9, d = ±3.
Numbers are 2, 5, 8 or 8, 5, 2.

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Common mistakes

  • Confusing the index: an uses n starting at 1, not 0. Always verify a1 gives the first term.
  • In GP, when r is negative, alternate terms change sign — keep track of signs carefully.
  • The formula S = a/(1-r) for infinite GP only applies when |r| < 1. If |r| ≥ 1, the series diverges.

Summary

APs have constant differences; GPs have constant ratios. Master the general term and sum formulas. The special sum formulas for natural numbers, squares, and cubes are frequently tested.

Practice Problems

15 questions with instant feedback.

Question 1 of 15Score 0

What is the 10th term of the AP: 2, 5, 8, 11, ...?