Answer
$a_8=19
\text{ and }
a_n=3+2n
$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find $a_8$ and $a_n$ in the given sequence
\begin{array}{l}\require{cancel}
5,7,9,...
,\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=7-5
\\\\
d=2
,\end{array}
Using $a_n=a_1+(n-1)d$ with $a_1=5$ and $d=2$ then
\begin{array}{l}\require{cancel}
a_n=5+(n-1)(2)
\\\\
a_n=5+2n-2
\\\\
a_n=3+2n
.\end{array}
With $n=8,$ then
\begin{array}{l}\require{cancel}
a_8=3+2(8)
\\\\
a_8=3+16
\\\\
a_8=19
.\end{array}
Hence, $
a_8=19
\text{ and }
a_n=3+2n
.$