Answer
$r=3^{2/3}$
$a_5=3^{11/3}$
The $n^{th}$ term of the geometric sequence is: $a_n=3 \cdot \left(3^{2/3}\right)^{n-1}$
Work Step by Step
RECALL:
(1) The common ratio of a geometric sequence is equal to the quotient of any two consecutive terms:
$r =\dfrac{a_n}{a_{n-1}}$
(2) The $n^{th}$ term of a geometric sequence is given by the formula:
$a_n = a\cdot r^{n-1}$
where
$a$ = first term
$r$ = common ratio
The sequence is said to be geometric.
Thus, we can proceed to solving for the common ratio:
$r=\dfrac{3^{5/3}}{3}$
Use the rule $\dfrac{a^m}{a^n} = a^{m-n}$ to obtain:
$r = 3^{5/3-1}
\\r=3^{5/3 - 3/3}
\\r=3^{2/3}$
The fifth term can be found by multiplying the common ratio to the fourth term.
The fourth term is $27$.
Thus, the fifth term is:
$a_5=27 \cdot \left(3^{2/3}\right)
\\a_5=3^3 \cdot 3^{2/3}$
Use the rule $a^m\cdot a^n = a^{m+n}$ to obtain:
$a_5=3^{3+2/3}
\\a_5=3^{9/3 + 2/3}
\\a_5=3^{11/3}$
With a first term of $3$ and a common ratio of $r=3^{2/3}$, the $n^{th}$ term of the geometric sequence is:
$a_n=a \cdot r^{n-1}
\\a_n=3 \cdot \left(3^{2/3}\right)^{n-1}$
