Question:
Which term of the progression $18,-12,8, \ldots$ is $\frac{512}{729} ?$
Solution:
Here, first term, $a=18$
and common ratio, $r=\frac{-2}{3}$
Let the $n^{\text {th }}$ term be $\frac{512}{729}$.
$\therefore a r^{n-1}=\frac{512}{729}$
$\Rightarrow(18)\left(\frac{-2}{3}\right)^{n-1}=\frac{512}{729}$
$\Rightarrow\left(\frac{-2}{3}\right)^{n-1}=\frac{512}{729} \times \frac{1}{18}=\frac{256}{6561}$
$\Rightarrow\left(\frac{-2}{3}\right)^{n-1}=\left(\frac{-2}{3}\right)^{8}$
$\Rightarrow n-1=8$
$\Rightarrow n=9$
Thus, the $9^{\text {th }}$ term of the given G.P. is $\frac{512}{729}$.