## Precalculus (10th Edition)

$a_n=2\cdot 3^{n-1}$ and $a_{5}=162.$
We know that if $a_1$ is the first term and $r$ is the common ratio, then the nth term is: $a_n=a_1\cdot r^{n-1}.$ Hence here $a_n=2\cdot 3^{n-1}$ and $a_{5}=2\cdot 3^{5-1}=162.$