Answer
$a_8=s+14p
\text{ and }
a_n=s+2np-2p$
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_4=s+6p, d=2p
,\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 $a_{4}=s+6p,$ $d=2p,$ and $n=4$ then
\begin{array}{l}\require{cancel}
a_4=a_1+(4-1)d
\\\\
s+6p=a_1+(3)2p
\\\\
s+6p=a_1+6p
\\\\
s+6p-6p=a_1
\\\\
s=a_1
\\\\
a_1=s
.\end{array}
Using $a_n=a_1+(n-1)d$ with $a_{1}=s,$ and $d=2p,$ then
\begin{array}{l}\require{cancel}
a_n=a_1+(n-1)d
\\\\
a_n=s+(n-1)2p
\\\\
a_n=s+2np-2p
.\end{array}
With $n=8,$ then
\begin{array}{l}\require{cancel}
a_8=s+2(8)p-2p
\\\\
a_8=s+16p-2p
\\\\
a_8=s+14p
.\end{array}
Hence, $
a_8=s+14p
\text{ and }
a_n=s+2np-2p
.$
