Answer
$a=\displaystyle \frac{81}{2},\quad a_{n}=\frac{81}{2}(\frac{2}{3})^{n-1}$
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: $a_{4}=12,\displaystyle \quad a_{7}=\frac{32}{9},$
$\displaystyle \frac{a_{7}}{a_{4}}=\frac{\frac{32}{9}}{12}=\frac{32\cdot 1}{9\cdot 12}=\frac{8}{9\cdot 3}=\frac{8}{27}$
Also, $\displaystyle \frac{a_{7}}{a_{4}}=\frac{ar^{6}}{ar^{3}}=r^{3}$
So, $r^{3}=\displaystyle \frac{8}{27}=(\frac{2}{3})^{3}$
$r=\displaystyle \frac{2}{3}$
$a_{4}=ar^{3}$ and $a_{4}=12$ leads to
$a(\displaystyle \frac{2}{3})^{3}=12$
$\displaystyle \frac{8}{27}a=12\qquad/\times\frac{27}{8}$
$a=\displaystyle \frac{12\cdot 27}{8}=\frac{3\cdot 27}{2}=\frac{81}{2}$
And, finally
$a_{n}=ar^{n-1}=\displaystyle \frac{81}{2}(\frac{2}{3})^{n-1}$
