CBSE Class 10 Mathematics Arithmetic Progressions Notes Set 09

Arithmetic Progression (A.P.)

An arithmetic progression is a sequence in which every term after the first is obtained from the previous term by adding the same fixed number, called the common difference d. Examples: 1, 2, 3, ..., 25 (d = 1) and 2, 7, 12, 17, ... (d = 5).

The nth Term

If the first term is a and the common difference is d, then the terms are a, a + d, a + 2d, ... and the nth term is a_n = a + (n − 1)d.

Sum of n Terms

Write the sum forwards and backwards and add: each pair of terms adds up to 2a + (n − 1)d, and there are n such pairs. So S_n = (n/2)[2a + (n − 1)d] = (n/2)(first term + last term).

Solved Examples on A.P.

1. Find the 20th term of 1, 4, 7, 10, ... Here a = 1, d = 3, so a₂₀ = 1 + 19 × 3 = 58.
2. Which term of 5, 8, 11, ... is 320? 5 + (n − 1) × 3 = 320 gives n − 1 = 105, so it is the 106th term.
3. Find x if 1, 3 + x, 17 − 4x are in A.P. The differences are equal: (3 + x) − 1 = (17 − 4x) − (3 + x), so 2 + x = 14 − 5x and x = 2 (the terms are 1, 5, 9).
4. The first term is −7 and d = 5. Then a₁₈ = −7 + 17 × 5 = 78 and the general term is a_n = −7 + 5(n − 1) = 5n − 12.
5. Find the sum of the first 30 terms of an A.P. whose second term is 2 and seventh term is 22. From a + d = 2 and a + 6d = 22: d = 4, a = −2. S₃₀ = 15[−4 + 29 × 4] = 15 × 112 = 1680.
6. The sum of 12 terms of an A.P. is 222 and the 12th term is 35. Then 222 = 6(a + 35), so a = 2, and 35 = 2 + 11d gives d = 3.
7. How many terms of 18, 15, 12, ... give a sum of 60? (n/2)[36 − 3(n − 1)] = 60 gives n² − 13n + 40 = 0, so n = 5 or n = 8. Both are valid because the 6th, 7th and 8th terms (3, 0, −3) add up to zero.
8. Three numbers in A.P. have sum −3 and product 8. Take a − d, a, a + d: 3a = −3 so a = −1; a(a² − d²) = 8 gives d² = 9, d = ±3. The numbers are −4, −1, 2.

Geometric Progression (G.P.)

A geometric progression is a sequence in which the ratio of every term to the one before it is the same number r, the common ratio. Examples: 2, 6, 18, 54, ... (r = 3) and 24, −12, 6, −3, ... (r = −1/2).

nth Term and Sum of a G.P.

The nth term is a_n = arⁿ⁻¹, where a is the first term.
For the sum S = a + ar + ... + arⁿ⁻¹, multiply by r and subtract: (1 − r)S_n = a(1 − rⁿ). Hence S_n = a(1 − rⁿ)/(1 − r) for r ≠ 1 (and S_n = na if r = 1).

Solved Example on G.P.

The first term of a G.P. is 3 and the sixth term is 96. Find the tenth term and the sum of the first ten terms. From 3r⁵ = 96: r⁵ = 32, so r = 2. The tenth term is 3 × 2⁹ = 1536. The sum is 3(2¹⁰ − 1)/(2 − 1) = 3 × 1023 = 3069.