Consider a sequence of numbers where the first term is $a_1$, the second term is $a_2$, the third term is $a_3$, and so on. The common ratio, or constant that is being multiplied to each term is $r$, and $a_n$ is the $n^{\rm th}$ term.

A pattern (formula) emerges. Thus:

$a_1=a_1$
$a_2=a_1r$
$a_3=a_2r=a_1r\cdot r=a_1r^2$
$a_4=a_3r=a_1r^2\cdot r=a_1r^3$
$a_5=a_4r=a_1r^3\cdot r=a_1r^4$
$\vdots$
$a_n=a_1r^{n−1}$

This is the formula for the $n^{\rm th}$ term of a geometric sequence.

Example 1: Consider the sequence of numbers: $1, ~3, ~9, ~27,$ $\ldots$ Suppose we want to find the $\rm 11^{th}$ term of the sequence.

$a_1=1$, $r=3$, and $n=11$. Using the formula for the $n^{\rm th}$ term, $a_n=a_1r^{n−1}$ gives:

$a_{11}=(1)(3)^{11−1}=(3)^{10} =59~049$

If we want to determine the sum of all the terms, then another pattern (formula) emerges. Without going into a formal proof, the sum of the first $n$ terms of a geometric sequence, called a geometric series, is given by:

$\mathrm S_n = \dfrac{a_1(1−r^n)}{1−r}$

Therefore, we need to know a few pieces of information. Namely, the first term $a_1$, how many terms are we summing $n$, and the common ratio $r$.

Example 2: Consider the sequence of numbers $2$, $6$, $18$, $54$, $\ldots$ Suppose we want to determine the sum of the first $8$ terms, or $\rm S_8$.

The sum of the first $8$ terms, $\rm S_8$, can be computed as:

$\mathrm S_n=\dfrac{a_1(1−r^n)}{1−r}$
$\rm S_8=\dfrac{2(1−3^8)}{1−3}$ $=\dfrac{2(1−6~561)}{−2}$ $=\dfrac{2(−6~560)}{−2}$ $=\dfrac{−13~120}{−2}=6~560$