Sequences
Find the nth term and the sum of an arithmetic or geometric sequence.
aₙ = a₁ + (n−1)d
Sₙ = n(a₁ + aₙ) / 2
For an arithmetic sequence the nth term is aₙ = a₁ + (n−1)d and the sum is Sₙ = n(a₁ + aₙ)/2. With a₁ = 2 and d = 3, the 5th term is a₅ = 2 + 4·3 = 14, and the sum of the first five terms is S₅ = 5(2 + 14)/2 = 40.
Arithmetic and geometric sequences
A sequence is an ordered list of numbers. In an arithmetic sequence you add a fixed amount — the common difference d — to get from one term to the next: 2, 5, 8, 11, … In a geometric sequence you multiply by a fixed factor — the common ratio r — instead: 2, 6, 18, 54, … This tool finds the nth term aₙ and the sum of the first n terms Sₙ for either kind.
nth term — arithmetic (left) and geometric (right)
Worked example
Take the arithmetic sequence with a₁ = 2 and d = 3, and find the 5th term and the sum of the first five terms.
- 1 Identify the first term and the step. Here a₁ = 2 and the common difference d = 3 (each term is 3 more than the last).
- 2 Apply the nth-term formula. aₙ = a₁ + (n−1)d, so a₅ = 2 + (5−1)·3 = 2 + 12 = 14.
- 3 Apply the sum formula. Sₙ = n(a₁ + aₙ)/2, so S₅ = 5·(2 + 14)/2 = 5·16/2 = 40.
- 4 For geometric, swap in the ratio. Use aₙ = a₁·rⁿ⁻¹ and Sₙ = a₁(rⁿ − 1)/(r − 1); for a₁ = 2, r = 3, n = 4 this gives a₄ = 54 and S₄ = 80.
Arithmetic vs geometric formulas
d is the common difference; r is the common ratio (r ≠ 1 in the geometric sum).
| Arithmetic | Geometric | |
|---|---|---|
| Next term from previous | add d | multiply by r |
| nth term aₙ | a₁ + (n−1)d | a₁·rⁿ⁻¹ |
| Sum Sₙ | n(a₁ + aₙ)/2 | a₁(rⁿ − 1)/(r − 1) |
| Example | 2, 5, 8, 11, … | 2, 6, 18, 54, … |
Difference vs ratio, and infinite sums
The common difference measures additive growth — every term sits a constant distance from its neighbour, so the terms trace a straight line. The common ratio measures multiplicative growth — each term is a constant multiple of the last, so the terms grow (or shrink) exponentially.
A geometric series can even be summed to infinity. When |r| < 1 the terms shrink toward zero fast enough that the total converges to S = a₁/(1 − r). For example, 1 + ½ + ¼ + ⅛ + … = 1/(1 − ½) = 2. When |r| ≥ 1 the terms do not shrink and the infinite sum diverges.