#### Answer

$r=\dfrac{2}{3}$
$a_5=\dfrac{112}{81}$
The $n^{th}$ term of the geometric sequence is: $a_n=7 \cdot \left(\dfrac{2}{3}\right)^{n-1}$

#### 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:
$r=\dfrac{\frac{14}{3}}{7}
\\r = \dfrac{14}{3} \cdot \dfrac{1}{7}
\\r=\dfrac{2}{3}$
The fifth terms can be found by multiplying the common ratio to the fourth term.
The fourth term is $\dfrac{56}{27}$.
Thus, the fifth term is:
$a_5=\dfrac{56}{27} \cdot \dfrac{2}{3}
\\a_5=\dfrac{112}{81}$
With a first term of $7$ and a common ratio of $r=\dfrac{2}{3}$, the $n^{th}$ term of the geometric sequence is:
$a_n=a \cdot r^{n-1}
\\a_n=7 \cdot \left(\dfrac{2}{3}\right)^{n-1}$