Answer
$a_n=6561\left(\frac{1}{3}\right)^{n-1}$
Work Step by Step
$a_6=27$ and $a_{12}=\frac{1}{27}$
$\frac{a_{12}}{a_6}=\frac{ar^{11}}{ar^5}=\frac{\frac{1}{27}}{27}=\frac{1}{729}=r^6$
$\implies r=\left(\frac{1}{729}\right)^{\frac{1}{6}}=\frac{1}{3}.$
$a_6=ar^5=a\left(\frac{1}{3}\right)^5=a\left(\frac{1}{243}\right)=27$
$\implies a=6561$
The $n^{th}$ term is therefore
$a_n=6561\left(\frac{1}{3}\right)^{n-1}$
