« Arithmetic Progressions 

By geometric progression of $m$ terms, we mean a finite sequence of the form
$$a, \; a r, \; a r^2, \; ..., \; a r^{m1} $$The real number $a$ is known as the first term of the geometric progression, and the real number $r$ is called the ratio of the geometric progression.
Example 1:
Consider the finite sequence of numbers
$$ 4, \; 8, \; 16, \; 32, \; 64, \; 128, \; 256, \; 512, \; 1024$$In this sequence, the ratio between successive terms is constant and equal to 2.
Here, we have: $a = 4$ and $r = 2$.
The $kth$ term of the geometric progression is equal to
$$ \color{blue}{a r^{k1}} $$The sum of the m terms of a geometric progression is equal to
$$ \color{blue} { S = \left\{ {\begin{array}{*{20}{c}} {a \cdot m \hspace{1.8cm} \text{if} \ r = 1 }\\ {\frac{{a  a{r^m}}}{{1  r}} \hspace{1cm} \text{if} \ r \ne 1} \end{array}} \right. } $$Example 2:
Consider the geometric sequence $1, \; \frac{1}{2}, \; \frac{1}{4}, \; \frac{1}{8},...$
Here we have: $a = 1$ and $r = \frac {1}{2}$.
The sum of the first $m$ terms is equal to
$$ \frac{a  a{r^m}}{1  r} = \frac{1  1 \cdot {( \frac{1}{2} )}^m}{1  \frac{1}{2}} = 2  {( \frac{1}{2} )}^{m  1} = 2  \frac{1}{2^{m  1}} $$This value gets very close to 2 if $m$ is very large.