#### Answer

$${a_n} = - 4{\left( { - 2} \right)^{n - 1}}{\text{ and }}{a_5} = - 64$$

#### Work Step by Step

$$\eqalign{
& {a_1} = - 4,\,\,\,\,r = - 2 \cr
& {\text{The general term of a geometric sequence is }}{a_n} = a{r^{n - 1}}.{\text{ with first term }}a{\text{ and }} \cr
& {\text{common ratio }}r. \cr
& {\text{substituting }}{a_1} = a = - 4\,\,\,\,r = - 2{\text{ into }}{a_n} = a{r^{n - 1}}{\text{ to obtain}} \cr
& {a_n} = {a_1}{r^{n - 1}} = - 4{\left( { - 2} \right)^{n - 1}} \cr
& \cr
& {\text{find }}{a_5},{\text{ substitute }}n = 5{\text{ into the general term formula}} \cr
& {a_5} = - 4{\left( { - 2} \right)^{5 - 1}} \cr
& {a_5} = - 4{\left( { - 2} \right)^4} \cr
& {a_5} = - 64 \cr
& {\text{then}} \cr
& {a_n} = - 4{\left( { - 2} \right)^{n - 1}}{\text{ and }}{a_5} = - 64 \cr} $$