Answer
$a=-\displaystyle \frac{1}{27},\quad a_{2}=\frac{1}{9}$
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: $a_{3}=-\displaystyle \frac{1}{3},\quad a_{6}=9,$
$\displaystyle \frac{a_{6}}{a_{3}}=-\frac{9}{\frac{1}{3}}=-27$
Also, $\displaystyle \frac{a_{6}}{a_{3}}=\frac{ar^{5}}{ar^{2}}=r^{3}$
So, $r^{3}=-27$
$r=-3$
$a_{3}=ar^{2}$ and $a_{3}=-\displaystyle \frac{1}{3}$ leads to
$a(-3)^{2}=-\displaystyle \frac{1}{3}$
$9a=-\displaystyle \frac{1}{3}$
$a=-\displaystyle \frac{1}{27}$
And, finally $a_{2}=ar=-\displaystyle \frac{1}{27}(-3)=\frac{1}{9}$
