Answer
$a_5=80
\text{ and }
a_n=5(-2)^{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_1=5
\\
r=-2
,\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 $a_n=a_1r^{n-1}$ with $a_1=5$ and $r=-2,$ then
\begin{array}{l}\require{cancel}
a_n=a_1r^{n-1}
\\\\
a_n=5(-2)^{n-1}
.\end{array}
With $n=5,$ then
\begin{array}{l}\require{cancel}
a_5=5(-2)^{5-1}
\\\\
a_5=5(-2)^{4}
\\\\
a_5=5(16)
\\\\
a_5=80
.\end{array}
Hence, $
a_5=80
\text{ and }
a_n=5(-2)^{n-1}
.$
