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Q.
Which term of the $G.P. 2, 1, \frac{1}{2}, \frac{1}{4}, ...$ is $\frac{1}{128}$?
Sequences and Series
Solution:
Here, $a = 2$ and common ratio $\left(r\right) = 1/2$.
Let the $n^{th }$ term be $\frac{1}{128 }$. Then,
$a_{n} = \frac{1}{128} $
$\Rightarrow ar^{n-1} = \frac{1}{128}$
$ \Rightarrow 2\left(\frac{1}{2}\right)^{n-1} = \frac{1}{128} $
$\Rightarrow \left(\frac{1}{2}\right)^{n-1} = \left(\frac{1}{2}\right)^{7}$
$ \Rightarrow n-2 = 7$
$ \Rightarrow n= 9$
Thus, $9^{th}$ term of the given $G.P$. is $\frac{1}{128}$.