Answer
$${a_n} = - {\left( { - 4} \right)^{n - 1}}{\text{ and }}{a_5} = - 256$$
Work Step by Step
$$\eqalign{
& {a_4} = 64,\,\,\,\,r = - 4 \cr
& {\text{The general term of a geometric sequence is }}{a_n} = a{r^{n - 1}}.{\text{ with first term }}a{\text{ and }} \cr
& {\text{common ratio }}r. \cr
& {\text{substituting }}{a_4} = 64\,\,\,\,r = - 4,\,\,\,n = 4{\text{ into }}{a_n} = a{r^{n - 1}}{\text{ to obtain}} \cr
& 64 = {a_1}{\left( { - 4} \right)^{4 - 1}} \cr
& 64 = {a_1}\left( { - 64} \right) \cr
& {a_1} = - 1 \cr
& {\text{find the general term }}{a_n} = {a_1}{r^{n - 1}} \cr
& {a_n} = - {\left( { - 4} \right)^{n - 1}} \cr
& \cr
& {\text{find }}{a_5},{\text{ substitute }}n = 5{\text{ into the general term formula}} \cr
& {a_5} = - {\left( { - 4} \right)^{5 - 1}} \cr
& {a_5} = - {\left( { - 4} \right)^4} \cr
& {a_5} = - 256 \cr
& {\text{then}} \cr
& {a_n} = - {\left( { - 4} \right)^{n - 1}}{\text{ and }}{a_5} = - 256 \cr} $$
