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  4. Which term of the following sequence (1/3), (1/9), (1/27)....... is (1/19683) ?

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Q. Which term of the following sequence $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}.......$ is $\frac{1}{19683} $ ?

Sequences and Series

Solution:

$a=\frac{1}{3}, r=\frac{1 / 9}{1 / 3}=\frac{1}{9} \times \frac{3}{1}=\frac{1}{3}$
Let $T_{n}=\frac{1}{19683}$
$ \Rightarrow a r^{n-1}=\frac{1}{19683}$
$\Rightarrow \frac{1}{3}\left(\frac{1}{3}\right)^{n-1}=\frac{1}{19683} $
$\Rightarrow \left(\frac{1}{3}\right)^{n}=\left(\frac{1}{3}\right)^{9} $
$\Rightarrow n=9$