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

Now that we know the value of $r$, we can substitute it into either equation  or  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.