Answer
$a_8=y-17;
\\a_n=(y+4)+(n-1)(-3)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find $a_8$ and $a_n$ in the given sequence \begin{array}{l}\require{cancel} a_2=y+1, d=-3 ,\end{array} use the formula for finding the $n$th term of an arithmetic sequence.
$\bf{\text{Solution Details:}}$
Using $a_n=a_1+(n-1)d$ with $a_{2}=y+1,$ $d=-3,$ and $n=2$ then
\begin{array}{l}\require{cancel}
a_n=a_1+(n-1)d
\\\\
a_2=a_1+(2-1)d
\\\\
a_2=a_1+(1)d
\\\\
a_2=a_1+d
\\\\
y+1=a_1+(-3)
\\\\
y+1=a_1-3
\\\\
y+1+3=a_1
\\\\
y+4=a_1
\\\\
a_1=y+4
.\end{array}
Using $a_n=a_1+(n-1)d$ with $a_{1}=y+4,$ and $d=-3,$ then
\begin{array}{l}\require{cancel}
a_n=a_1+(n-1)d
\\\\
a_n=y+4+(n-1)(-3)
.\end{array}
With $n=8,$ then \begin{array}{l}\require{cancel}
a_8=y+4+(8-1)(-3)
\\\\
a_8=y+4+(7)(-3)
\\\\
a_8=y+4-21
\\\\
a_8=y-17
.\end{array}
Hence, $
a_8=y-17 \text{ and } a_n=y+4+(n-1)(-3)
.$