Answer
a. $a_{n} = \sqrt 3(\sqrt 3)^{n-1}$
b. Fourth term: 9
Work Step by Step
a. $a_{n} = a(r)^{n-1}$ is the formula for the nth term of a geometric sequence. In this case a = $\sqrt{3}$ and r=$\sqrt{3}$. Plug these values into the formula to get a formula for the nth term of this sequence.
b. To find the fourth term plug in n = 4 into the sequence: $a_{n} = \sqrt{3}(\sqrt{3})^{4-1} = \sqrt{3}(\sqrt{3})^{3} = 9$