Answer
$a_n=81\left(\frac{1}{3}\right)^{n-1}$
Work Step by Step
$81, 27, 9, 3, 1, \frac{1}{3} , . . . $
$\frac{a_2}{a_1}=\frac{27}{81}=\frac{1}{3}=r$
$ra_2=\frac{1}{3}\times27=9=a_3$
$ra_3=\frac{1}{3}\times9=3=a_4$
and so on. The sequence is geometric with ratio $r=\frac{1}{3}$. The $n^{th}$ term is given by
$a_n=81\left(\frac{1}{3}\right)^{n-1}$