Answer
$a_n=1\cdot \left(-\frac{1}{3}\right)^{n-1}$ and $a_{5}=\frac{1}{81}.$
Work Step by Step
We know that if $a_1$ is the first term and $r$ is the common ratio, then the nth term is: $a_n=a_1\cdot r^{n-1}.$
Hence here $a_n=1\cdot \left(-\frac{1}{3}\right)^{n-1}$ and $a_{5}=1\cdot \left(-\frac{1}{3}\right)^{5-1}=\frac{1}{81}.$