In an arithmetic sequence u5 = 23 and u12 = 58. Find the sum of the first 20 terms.
- u12 − u5 = 7d, so 7d = 58 − 23 = 35 and d = 5.
- u5 = u1 + 4d, so u1 = 23 − 4 × 5 = 3.
- S20 = (20/2) × (2 × 3 + 19 × 5) = 10 × (6 + 95) = 10 × 101.
Answer: S20 = 1010
Arithmetic sequences and series
Maths: Analysis and Approaches · Number and algebra. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.
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The idea
An arithmetic sequence is a list of numbers that goes up or down by the same amount each time. That fixed amount is the common difference, d. It can be negative, so the terms can get smaller.
To reach the nth term you start at the first term u1 and add d a total of (n − 1) times, not n times. An arithmetic series is the sum of the terms. Sigma notation is a short way to write that sum: a linear expression such as 3k + 2 always gives an arithmetic series.
Real data is rarely perfectly arithmetic. In a model you use an average difference and treat the answer as an estimate.
Rules to remember
- nth term = first term + (n − 1) × common difference (un = u1 + (n − 1)d)
- Common difference: d = u2 − u1 = u3 − u2, the same for every pair of neighbours
- Sum of the first n terms: Sn = (n/2) × (2u1 + (n − 1)d)
- If you know the last term: Sn = (n/2) × (u1 + un)
- Two terms given: um − uk = (m − k)d, which gives d straight away
- A single term from sums: un = Sn − Sn − 1
- Σ from k = 1 to n has n terms; put k = 1 to get u1 and k = n to get the last term
Common mistake
Using n × d instead of (n − 1) × d in the nth term. The first term has had no differences added, so the 20th term has only 19 of them.
Worked example
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