Answer
The common ratio is $3^{\frac{2}{3}}$.
The 5th term is $3^\frac{11}{3}$.
$a_{n} = 3(3^{\frac{2}{3}})^{n-1}$
Work Step by Step
The terms given are 3 $3^{\frac{5}{3}}$ $3^{\frac{7}{3}}$ 27.
Since each term is multiplied by $3^{\frac{2}{3}}$ to get to the next, the common ratio is $3^{\frac{2}{3}}$ The fifth term = the fourth term $\times$ common ration = 27 $\times $ $3^{\frac{2}{3}}$$ = 3^\frac{11}{3}$. The formula for the nth term of a geometric sequence is written in the form $a_{n} = a(r)^{n-1}$ since the first term (a) is 3 and the common ratio (r) is $3^{\frac{2}{3}}$.
Plug these into the formula to get $a_{n} = 3(3^{\frac{2}{3}})^{n-1}$
