Answer
$a_{n}=\frac{1}{7}(n+2)$
$a_{10}=\frac{12}{7}$
Work Step by Step
The $n$th term is given by
$a_{n}=a_{1}+(n-1)d$
where $a_{1}$ is the first term and $d$ is the common difference.
Substituting $\frac{3}{7}$ for $a_{1}$ and $\frac{1}{7}$ for $d$, we get
$a_{n}=\frac{3}{7}+(n-1)\frac{1}{7}$
Simplifying, we have
$a_{n}=\frac{1}{7}(n+2)$
Substituting $n=10$, we obtain
$a_{10}=\frac{1}{7}(10+2)=\frac{12}{7}$
