#### Answer

$a = -\frac{9}{32}$
The $n^{th}$ term of the given sequence is: $a_n = -\frac{9}{32} \cdot (-8)^{n-1}$

#### Work Step by Step

To find the first and nth terms, the value of the common ratio $r$ is needed.
Note that the next term of a geometric sequence can be found by multiplying the common ratio $r$ to the current term.
The known terms are:
$a_3=-18
\\a_6 =9216$
Note that using the third term as reference point, the sixth term can be found by multiplying the common ratio $r$ three times to the value of the third term.
Thus,
$a_6=a_3 \cdot r \cdot r \cdot r
\\a_6 = a_3 \cdot r^3$
Since $a_3=-18$ and $a_6=9216$, substituting these values gives:
$a_6=a_3 \cdot r^3
\\9216 = -18 \cdot r^3
\\\dfrac{9216}{-18} = \dfrac{-18r^3}{-18}
\\-512 = r^3
\\(-8)^3 = r^3$
Take the cube root of both sides to obtain:
$-8=r$
With $r=-8$, the first term can be computed by dividing $a_3$ by the common ratio twice to obtain:
$a = a_3 \div (r \times r)
\\a = a_3 \div r^2
\\a=-18 \div (-8)^2
\\a= -18 \div 64
\\a = -\frac{9}{32}$
RECALL:
The $n^{th}$ term $a_n$ of a geometric sequence can be found using the formula
$a_n = a \cdot r^{n-1}$
where
$a$ = first term
$r$ = common ratio
Thus, the $n^{th}$ term of the given sequence is:
$a_n = -\frac{9}{32} \cdot (-8)^{n-1}$