Answer
$a_8=-31
\text{ and }
a_n=1-4n$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find $a_8$ and $a_n$ in the given sequence
\begin{array}{l}\require{cancel}
-3,-7,-11,...
,\end{array}
use the formula for finding the $n$th term of an arithmetic sequence.
$\bf{\text{Solution Details:}}$
Since the common difference, $d,$ is the difference between a term and the term preceeding it, then
\begin{array}{l}\require{cancel}
d=a_2-a_1
\\\\
d=-7-(-3)
\\\\
d=-7+3
\\\\
d=-4
.\end{array}
Using $a_n=a_1+(n-1)d$ with $a_1=-3$ and $d=-4$ then
\begin{array}{l}\require{cancel}
a_n=-3+(n-1)(-4)
\\\\
a_n=-3-4n+4
\\\\
a_n=1-4n
.\end{array}
With $n=8,$ then
\begin{array}{l}\require{cancel}
a_8=1-4(8)
\\\\
a_8=1-32
\\\\
a_8=-31
.\end{array}
Hence, $
a_8=-31
\text{ and }
a_n=1-4n
.$
