Answer
$a_n=(\sqrt{3})^{n}$
$a_4=9$
Work Step by Step
We are given:
$a=\sqrt{3}, r=\sqrt{3}$
We know that a geometric sequence has the form:
$a_{n}=ar^{n-1}$
Thus:
$a_n=\sqrt{3}(\sqrt{3})^{n-1}=(\sqrt{3})^{n}$
And so:
$a_4=(\sqrt{3})^4=9$