Which term of the progression 18, −12, 8, ... is

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 $\frac{512}{729}$.

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