Answer
$a_8=29 \text{ and } a_n=-3n+53$
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_{15}=8, a_{16}=5 ,\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_{16}-a_{15} \\\\ d=5-8 \\\\ d=-3 .\end{array}
Using $a_n=a_1+(n-1)d$ with $a_{15}=8$ $d=-3,$ and $n=15,$ then \begin{array}{l}\require{cancel} a_n=a_1+(n-1)d \\\\ a_{15}=a_1+(15-1)d \\\\ 8=a_1+(14)(-3) \\\\ 8=a_1-42 \\\\ 8+42=a_1 \\\\ a_1=50 .\end{array}
Using $a_n=a_1+(n-1)d$ with $a_1=50$ and $d=-3,$ then \begin{array}{l}\require{cancel}
a_n=50+(n-1)(-3)
\\\\
a_n=50-3n+3
\\\\
a_n=-3n+53
.\end{array}
With $n=8,$ then \begin{array}{l}\require{cancel}
a_8=-3(8)+53
\\\\
a_8=-24+53
\\\\
a_8=29
.\end{array}
Hence, $
a_8=29 \text{ and } a_n=-3n+53
.$
