Answer
$${a_n} = 4{\left( 3 \right)^{n - 1}}{\text{ and }}{a_5} = 324$$
Work Step by Step
$$\eqalign{
& {a_1} = 4,\,\,\,\,r = 3 \cr
& {\text{The general term of a geometric sequence is }}{a_n} = a{r^{n - 1}}.{\text{ with first term }}a{\text{ and }} \cr
& {\text{common ratio }}r. \cr
& {\text{substituting }}{a_1} = a = 4,\,\,\,\,r = 3{\text{ into }}{a_n} = a{r^{n - 1}}{\text{ to obtain}} \cr
& {a_n} = {a_1}{r^{n - 1}} = 4{\left( 3 \right)^{n - 1}} \cr
& {a_n} = 4{\left( 3 \right)^{n - 1}} \cr
& \cr
& {\text{find }}{a_5},{\text{ substitute }}n = 5{\text{ into the general term formula}} \cr
& {a_5} = 4{\left( 3 \right)^{5 - 1}} \cr
& {a_5} = 324 \cr
& {\text{then}} \cr
& {a_n} = 4{\left( 3 \right)^{n - 1}}{\text{ and }}{a_5} = 324 \cr} $$
