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Arithmetic sequences and series

In an arithmetic sequence you add the same number, the common difference d, each time. An arithmetic series is the sum of its terms.

nth term

First term a, common difference d: u_n = a + (n - 1)d.

Sum of n terms

S_n = (n/2)(2a + (n - 1)d), or S_n = (n/2)(a + l) where l is the last term.

The sum of the first n natural numbers is 1 + 2 + ... + n = n(n + 1)/2.

Proof of the sum formula

Write S_n = a + (a + d) + ... + (a + (n - 1)d), then write it again in reverse order. Adding the two lines, each pair adds to 2a + (n - 1)d and there are n pairs:

2S_n = n(2a + (n - 1)d), so S_n = (n/2)(2a + (n - 1)d). You must know this proof.

Worked example

Find the sum of the first 20 terms of 5 + 8 + 11 + ...

  1. a = 5, d = 3, n = 20
  2. S_20 = 10(2 x 5 + 19 x 3)
  3. = 10(10 + 57)

Answer: S_20 = 670

Key idea: u_n = a + (n - 1)d; S_n = (n/2)(2a + (n - 1)d) = (n/2)(a + l); 1 + 2 + ... + n = n(n + 1)/2.

Check yourself

  1. Find the 20th term of 7, 11, 15, ...
    Show answer

    83. u_20 = 7 + 19 x 4 = 83.

  2. Find the sum of the first 30 terms of 3 + 7 + 11 + ...
    Show answer

    1830. S_30 = 15(6 + 116) = 1830.

  3. Find 1 + 2 + 3 + ... + 100.
    Show answer

    5050. n(n + 1)/2 = 100 x 101 / 2 = 5050.

  4. An arithmetic sequence has u_3 = 11 and u_8 = 31. Find d.
    Show answer

    4. From the 3rd to the 8th term there are 5 steps: 5d = 20, so d = 4.

  5. Find the least number of terms of 2 + 5 + 8 + ... needed for the sum to exceed 500.
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    19. S_n = (n/2)(4 + 3(n - 1)) = n(3n + 1)/2. S_18 = 495 and S_19 = 551, so 19 terms are needed.

  6. Find the sum of the odd numbers 1 + 3 + 5 + ... + 99.
    Show answer

    2500. There are 50 terms, so S = 25 x (1 + 99) = 2500.

  7. In the proof of the sum formula, why is the series written out a second time in reverse?
    • So the common difference becomes -d
    • To find the last term
    • So each pair of terms adds to 2a + (n - 1)d
    • So the terms cancel to zero
    Show answer

    So each pair of terms adds to 2a + (n - 1)d. Lining up first with last, second with second last, and so on, gives n equal pairs.

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Last updated 7 November 2026. Spotted a mistake? Tell us.