Answer
$a_n=3 (\dfrac{1}{3})^{n-1}$
Work Step by Step
The $n^{th}$ term of the sequence is given by the formula:
$ a_n=a_1r^{n-1}$
where $r$=common ratio and $a_1$= the first term
The common ratio of a geometric sequence is equal to the quotient of any term and the term before it:
$ \ r = \dfrac{a_n}{a_{n-1}}$ or, $r=\dfrac{a_2}{a_1}$
Here, we have: $a_3 = a_1 (r)(r); \\a_3 = a_1 \ (1/3)^2 \ \\ \dfrac{1}{3}=\dfrac{a_1}{9} \implies a_1=3$
Therefore, the $n^{th}$ term of the geometric sequence is: $a_n=3 (\dfrac{1}{3})^{n-1}$
