Answer
$a_5=-108 \text{ and } a_n=-\dfrac{4}{3}\cdot3^{n-1}$
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} a_2=-4 \\ r=3 ,\end{array} use the formula for finding the $n^{th}$ term of a geometric sequence.
$\bf{\text{Solution Details:}}$
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_2=a_1r^{2-1} \\\\ a_2=a_1r^{1} \\\\ a_2=a_1r .\end{array}
Subtituting the given values, $a_2=-4$ and $r=3,$ the equation above results to \begin{array}{l}\require{cancel} -4=a_1(3) \\\\ -\dfrac{4}{3}=a_1 \\\\ a_1=-\dfrac{4}{3} .\end{array}
With $a_1=-\dfrac{4}{3}$ and $r=3,$ then \begin{array}{l}\require{cancel} a_n=a_1r^{n-1} \\\\
a_n=-\dfrac{4}{3}\cdot3^{n-1}
.\end{array}
With $n=5,$ the equation above bcomes \begin{array}{l}\require{cancel}
a_5=-\dfrac{4}{3}\cdot3^{5-1}
\\\\
a_5=-\dfrac{4}{3}\cdot3^{4}
\\\\
a_5=-\dfrac{4}{3}\cdot81
\\\\
a_5=-4(27)
\\\\
a_5=-108
.\end{array}
Hence, $
a_5=-108 \text{ and } a_n=-\dfrac{4}{3}\cdot3^{n-1}
.$
