Answer
$a=-384,\displaystyle \qquad a_{n}=-384(-\frac{3}{8})^{n-1}$
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 $\quad a_{n}=ar^{n-1}$
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Given: $a_{3}=-54,\displaystyle \quad a_{6}=\frac{729}{256}$
$\displaystyle \frac{a_{6}}{a_{3}}=\frac{\frac{729}{256}}{-54}=\frac{729}{-54\cdot 256}=-\frac{27}{512}$
Also, $\displaystyle \frac{a_{6}}{a_{3}}=\frac{ar^{5}}{ar^{2}}=r^{3}$
So, $r^{3}=-\displaystyle \frac{27}{512}=( -\frac{3}{8} )^{3}$
$r=-\displaystyle \frac{3}{8}$
$a_{3}=ar^{2}$ and $a_{3}=-54$ leads to
$a(-\displaystyle \frac{3}{8})^{2}=-54$
$\displaystyle \frac{9}{64}a=-54\qquad/\times\frac{64}{9}$
$a=\displaystyle \frac{-54\cdot 64}{9}=-6\cdot 64=-384$
And, finally
$a_{n}=ar^{n-1}=-384(-\displaystyle \frac{3}{8})^{n-1}$
