Answer
$a=-\displaystyle \frac{9}{32},\qquad a_{n}=-\frac{9}{32}(-8)^{n-1}$
Work Step by Step
A geometric sequence is a sequence whose terms are obtained by multiplying each term by the same fixed constant $r$ to get the next term.
A geometric sequence has the form
$a, ar, ar^{2}, ar^{3}, \ldots$
The number $a$ is the first term of the sequence, and the number $r $is the common ratio.
The nth term of the sequence is $\quad a_{n}=ar^{n-1}$
--------------
Given: $a_{3}=-18,\quad a_{6}=9216,$
$\displaystyle \frac{a_{6}}{a_{3}}=\frac{9216}{-18}=-512$
Also, $\displaystyle \frac{a_{6}}{a_{3}}=\frac{ar^{5}}{ar^{2}}=r^{3}$
So, $r^{3}=-512=( -8 )^{3}$
$r=-8$
$a_{3}=ar^{2}$ and $a_{3}=-18$ leads to
$a(-8)^{2}=-18$
$64a=-18\displaystyle \qquad/\times\frac{1}{64}$
$a=\displaystyle \frac{-18}{64}=-\frac{9}{32}$
And, finally
$a_{n}=ar^{n-1}=-\displaystyle \frac{9}{32}(-8)^{n-1}$
