A series where the ratio of any term (except the first) to the previous term is constant is called a geometric series or geometric progression which we abbreviate as G.P. 

The first term of a G.P. is denoted $a$. The ratio of any term to the preceding term is the common ratio $r$.

Thus $3 + 6 + 12 + 24 + 48$ is a G.P. with five terms, the first term is $a = 3$ and the common ratio is $r = 2$.

In general for a G.P. with common ratio $r$ we have

the first term is $a_1 = a$ 

the second term is $a_2=r\times a_1=r\times a = ar^1$

the third term is $a_3 = r \times a_2 = r \times ar^1 = ar^2$

the fourth term is $a_4 = r \times a_3 = r \times ar^2 = ar^3$

et cetera and in general the $n$ th term is $a_n=ar^{n\mathbin{-}1}$

thus in general a G.P. of $n$ terms is represented as $$a + ar + ar^2 + ar^3 + \cdots + ar^{n\mathbin{-}1}$$

Note the last or $n$ th term is $ar^{n \mathbin{-}1}$ not $ar^n$. If $r$ is positive all terms have the same sign as $a$ but if $r$ is negative then the terms alternate in sign

Consider the following G.P.

$$3+6+12+24+48$$

The number 6 is the geometric mean between 3 and 12 because 3, 6 and 12 are in geometric progression. Also 6 and 12 are two geometric means between 3 and 24 and 6, 12, 24 are three geometric means between 3 and 48. In general there may be any number of geometric means between any two numbers.

Exercises:

1. Find the tenth term of the G.P. with $a_5=12$ and $a_8=324$.

2. Find the tenth term of the G.P. with $a_1 + a_2 = 4$ and $a_2 + a_3 = 12$

3. Show that the geometric mean of $a$ and $b$ is $\pm\sqrt{ab}$

3. Insert 3 geometric means between 2 and 162.