Answer
$${a_n} = {\left( { - 3} \right)^n}\,\,\,{\text{and }}{a_5} = - 243$$
Work Step by Step
$$\eqalign{
& {a_4} = 81,\,\,\,\,r = - 3 \cr
& {\text{The general term of a geometric sequence is }}{a_n} = a{r^{n - 1}}.{\text{ with first term }}a{\text{ and }} \cr
& {\text{common ratio }}r. \cr
& {\text{substituting }}{a_4} = 81\,\,\,\,r = - 3,\,\,\,n = 4{\text{ into }}{a_n} = a{r^{n - 1}}{\text{ to obtain}} \cr
& 81 = {a_1}{\left( { - 3} \right)^{4 - 1}} \cr
& 81 = {a_1}\left( { - 27} \right) \cr
& {a_1} = - 3 \cr
& {\text{find the general term }}{a_n} = {a_1}{r^{n - 1}} \cr
& {a_n} = - 3{\left( { - 3} \right)^{n - 1}} \cr
& {a_n} = {\left( { - 3} \right)^n} \cr
& {\text{find }}{a_5},{\text{ substitute }}n = 5{\text{ into the general term formula}} \cr
& {a_5} = {\left( { - 3} \right)^5} \cr
& {a_5} = - 243 \cr
& \cr
& {\text{then}} \cr
& {a_n} = {\left( { - 3} \right)^n}\,\,\,{\text{and }}{a_5} = - 243 \cr} $$