How to calculate an arithmetic sequence — step by step
An arithmetic sequence is a list of numbers with a constant difference d between terms. The formula aₙ = a₁ + (n−1)·d gives any term, and Sₙ = n/2 · (2a₁ + (n−1)d) gives the sum. Worked example: a₁ = 3, d = 5, n = 10 → aₙ = 48 and Sₙ = 255. Suitable for Grade 10 / Year 11.
Quick answer
In an arithmetic sequence every term grows by the same difference d. The nth term is aₙ = a₁ + (n−1)·d. For a₁ = 3, d = 5, n = 10 that gives 3 + 9·5 = 48. The sum of the first n terms is Sₙ = n/2 · (2a₁ + (n−1)d) = 255.
At a glance
| Example | a₁ = 3, d = 5, n = 10 |
|---|---|
| Method | nth-term formula and sum formula |
| Formula | aₙ = a₁ + (n−1)·d |
| nth term | a₁₀ = 48 |
| Sum | S₁₀ = 255 |
| Grade level | Grade 10 (ages 15–16) |
Worked example: a₁ = 3, d = 5, n = 10
We want the 10th term and the sum of the first 10 terms of the sequence 3, 8, 13, 18, …
The steps for an arithmetic sequence
These steps work for any arithmetic sequence with first term a₁ and difference d.
Step 1 · Formula
aₙ = a₁ + (n − 1) · dGeneral formula for the nth term.Step 2 · Substitute
a₁₀ = 3 + (10 − 1) · 5Substituting a₁ = 3, d = 5 and n = 10.Step 3 · nth term
3 + 9 · 5 = 48The 10th term of the sequence is 48.Step 4 · Sum
S₁₀ = 10/2 · (2·3 + 9·5)Sum formula with the same values.Step 5 · Check
5 · 51 = 255(3+48)/2 · 10 = 255 confirms the result.
Why the formulas work
Because every term is exactly d larger than the one before, exactly (n−1) steps accumulate from the start a₁ to the nth term — hence aₙ = a₁ + (n−1)·d. The sum formula follows from the famous Gauss trick: pair the first term with the last, the second with the second-to-last, and so on, and every pair has the same sum a₁ + aₙ. With n terms there are n/2 such pairs, so Sₙ = n/2 · (a₁ + aₙ) = n/2 · (2a₁ + (n−1)d).