Answer
$a_5=-324
\text{ and }
a_n=-4(3)^{n-1}
$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find $a_5$ and $a_n$ for the given geometric sequence described as
\begin{array}{l}\require{cancel}
-4,-12,-36,-108,...
,\end{array}
use the formula for finding the $n^{th}$ term of a geometric sequence.
$\bf{\text{Solution Details:}}$
The common ratio, $r,$ is the quotient of a term and the term preceeding it. Hence,
\begin{array}{l}\require{cancel}
r=\dfrac{a_2}{a_1}
\\\\
r=\dfrac{-12}{-4}
\\\\
r=3
.\end{array}
Using the formula for finding the $n^{th}$ term of a geometric sequence, which is given by $a_n=a_1r^{n-1},$ then
\begin{array}{l}\require{cancel}
a_n=a_1r^{n-1}
\\\\
a_n=-4(3)^{n-1}
.\end{array}
With $n=5,$ the equation above bcomes
\begin{array}{l}\require{cancel}
a_5=-4(3)^{5-1}
\\\\
a_5=-4(3)^{4}
\\\\
a_5=-4(81)
\\\\
a_5=-324
.\end{array}
Hence, $
a_5=-324
\text{ and }
a_n=-4(3)^{n-1}
.$