Geometric Sequence: nth Term and Sum

From the first term and the common ratio, finds the nth term ar^(n−1) of a geometric sequence and the sum of the first n terms.

Explanation

In a geometric sequence each term is the previous one multiplied by a fixed number, the common ratio. The first term, the ratio and a term number give you both the nth term and the sum up to it.

a_n = a r^{n-1}

S_n = \dfrac{a(1 - r^n)}{1 - r} \quad (r \neq 1)

$a$ — the first term
$r$ — the common ratio, the multiplier applied at each step
$n$ — which term you want, an integer of 1 or more
$a_n$ — the nth term
$S_n$ — the sum of the first $n$ terms

The exponent is $n - 1$ , not $n$ . At the first term $r^0 = 1$ , which hands back the first term unchanged, exactly as it should.

Example

Take $a = 2$ , $r = 3$ and $n = 5$ . The sequence is 2, 6, 18, 54, 162.

a_5 = 2 \times 3^{5-1} = 2 \times 81 = 162

S_5 = \dfrac{2 \times (1 - 3^5)}{1 - 3} = \dfrac{2 \times (-242)}{-2} = 242

The fifth term is 162 and the sum is 242, which you can confirm by adding $2 + 6 + 18 + 54 + 162$ by hand.

Notes

$n$ must be a whole number of 1 or more.
The sum formula breaks at $r = 1$ , where the denominator would be zero. Every term is then just the first term, so the sum is simply $S_n = an$ . The calculator handles that case separately.
A negative ratio flips the sign of every other term.
When $|r| < 1$ the terms shrink towards zero, and letting $n$ grow without bound gives the convergent infinite series $\dfrac{a}{1 - r}$ .
With a ratio above 1 the nth term grows extremely fast, so large values of $n$ run into very big numbers quickly.