Find the sum of the first 15 terms of the series 4 + 9 + 14 + ...
- a = 4 and d = 9 − 4 = 5, with n = 15.
- S15 = (15/2)[2 × 4 + (15 − 1) × 5] = (15/2)(8 + 70).
- = (15/2) × 78 = 15 × 39 = 585.
Answer: S15 = 585
Arithmetic progressions
Maths · Sequences and series. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.
The idea
An arithmetic progression (A.P.) is a sequence in which each term is found by adding the same number to the term before it. That number is the common difference d, and the first term is written a (or a1).
Once you know a and d you know everything: any term, and the sum of any number of terms. So most questions come down to finding a and d first, often from two pieces of information that give two equations.
An arithmetic mean (A.M.) is a number placed between two numbers so that all of them are in A.P. If a, b, c are in A.P., then 2b = a + c.
Rules to remember
- common difference = any term − the term before it (d = an − an−1)
- nth term: an = a + (n − 1)d
- A.M. between a and b: A = (a + b)/2
- n arithmetic means between a and b: d = (b − a)/(n + 1), then the means are a + d, a + 2d, ...
- sum of n terms: Sn = (n/2)[2a + (n − 1)d] = (n/2)(a + an)
- 1 + 2 + 3 + ... + n = n(n + 1)/2
Common mistake
Students use nd in place of (n − 1)d. The first term has no d added, so the 20th term is a + 19d, not a + 20d.
Worked example
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