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 $n^{th}$ term of a geometric sequence. 

Example 1: Consider the sequence of numbers: $2$, $6$, $18$, $54$, $\ldots$
Suppose we want to find the $\rm 10^{th}$ term of the sequence. 

$a_1 = 2$, $r = 3$ and $n = 10$.

Using the formula for the $n^{\rm th}$ term, $a_n = a_1r^{n−1}$ gives: $a_{10} = (2)(3)10−1$ $= (2)(3)9$ $= (2)(19~683)$ $= 39~366$ 

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 previous sequence of numbers $2$, $6$, $18$, $54$, $\ldots$
Suppose we want to determine the sum of the first $10$ terms, or $\mathrm S_{10}$. 

The sum of the first $10$ terms, $\mathrm S_{10}$, can be computed as:

$\mathrm S_n = \dfrac{a_1(1−r^n)}{ 1−r}$ 
$\mathrm S_{10} = \dfrac{2(1 − 3^{10})}{1-3}$
$\mathrm S_{10} = \dfrac{2(1 − 59~049)}{-2}$
$\mathrm S_{10} = 
\dfrac{2(−59 048)}{-2}$
$\mathrm S_{10} = \dfrac{−118 ~096}{-2}$
$\mathrm S_{10} = 59~ 048$

If the common ratio $r$ is such that $−1 \leq r \leq 1$, then we can compute the sum of an infinite sequence of numbers, or an infinite series. The formula becomes much simpler and can be written as: 

$\mathrm S_{\infty} = \dfrac{a_1}{1−r}$

where $\rm S_{\infty}$ is the infinite sum of infinite terms, $a_1$ is the first term, and $r$ is the common ratio.

Example 3: Consider the sequence of numbers given by: $4$, $2$, $1$, $2$ , $\ldots$
We can $1$ continue this pattern and add infinitely many terms only because $r = \dfrac{1}{2}$, which is greater than $-1$ and less than $1$. Therefore:

$\mathrm S_{\infty} = \dfrac{a_1}{1-r} = \dfrac{4}{1 - \frac{1}{2}} = \dfrac{4}{\frac{1}{2}}=8$