The numbers of a sequence are called terms where the $n$th term of a sequence is denoted by the symbol $t_n$.

The Rule

A rule can be used which enables any term of a geometric sequence where we know the value of $a$ and $r$ to be found.

$t_n=ar^{n-1}$

Where $t_n$ is the $n$th term
$a$ is the first term
$r$ is the common ratio

Example 1

Find the 13th term of the geometric sequence 9, 27, 81, 243, 729, …

$a=9$
$n=13$
$r=\dfrac {27}{9}$
$r=3$

Use the rule:

$t_n=ar^{n-1}$
$t_{13}=9\times 3^{13-1}$
$t_{13}=9\times 531,441$
$t_{13}=4,782,969$

The value of the 13th term is 4 782 969.

Example 2

The 3rd term of a geometric sequence is 32 and the 6th is 2 048. Find the 12th term of this sequence.

We know that $t_3=32$ and that $t_n=ar^{n-1}$.
$t_3=ar^2$
$t_3=8$

We know that $t_6=2 048$ and that $t_n=ar^{n-1}$.
$t_6=ar^5$
$t_6=2,048$

We have two different equations so we need to solve them simultaneously.

[1]$a \times$$r^2=8$
[2]$a \times r^5=2,048$
[2] $\div$ [1]    $\dfrac {a \times r^5}{a \times r^2}$ $=\dfrac {2 048}{32}$
$r^3=64$
$r=4$

Now that we know the value of $r$, we can substitute it into either equation [1] or [2] to find the value of $a$.

$a\times 4=8$
$a=2$

Using the rule, we can now calculate the value of $t_{12}$

$t_n=ar^{n-1}$
$t_{12}=2\times 4^{12-1}$
$t_{12}=2\times 4^{11}$
$t_{12}=8,388,608$

The 12th term in the sequence is 8 388 608.