Answer
$a_n = 3 \cdot (\frac{1}{3})^{n-1}$
Work Step by Step
RECALL:
(1) The $n^{th}$ term $a_n$ of a geometric sequence is given by the formula
$a_n=a_1 \cdot r^{n-1}$
where
$a_1$ = first term
$r$ = common ratio
(2) To find the next term in a geometric sequence, the common ratio $r$ is multiplied by the current term.
Part (2) above implies that if the value of $a_3$ is known, the common ratio must be multiplied by $a_3$ three times to find the value of $a_6$.
Thus,
$a_6 = a_3 \cdot r \cdot r \cdot r
\\a_6 = a_3 \cdot r^3$
Substitute the given values of $a_3$ and $a_6$ into the equation above to obtain:
$a_6 = a_3 \cdot r^3
\\\frac{1}{81} = \frac{1}{3} \cdot r^3$
Multiply by $3$ on both sides to obtain:
$3 \cdot \frac{1}{81} = \frac{1}{3} \cdot r^3 \cdot 3
\\\frac{1}{27} = r^3
\\(\frac{1}{3})^3=r^3$
Take the cube root of both sides to obtain:
$\sqrt[3]{(\frac{1}{3})^3} = \sqrt[3]{r^3}
\\\frac{1}{3} = r$
Note that:
$a_3=a_1 \cdot r \cdot r
\\a_3 = a_1 \cdot r^2$
Substitute the known values of $a_3$ and $r$ into the equation above to obtain:
$a_3 = a_1 \cdot r^2
\\\frac{1}{3} = a_1 \cdot (\frac{1}{3})^2
\\\frac{1}{3} = a_1 \cdot \frac{1}{9}$
Multiply by $9$ on both sides of the equation to obtain:
$9(\frac{1}{3}) = a_1 \cdot \frac{1}{9} \cdot 9
\\3 = a_1$
Thus, the $n^{th}$ term of the geometric sequence is:
$a_n=a_1 \cdot r^{n-1}
\\a_n = 3 \cdot (\frac{1}{3})^{n-1}$
