How many terms of the G.P. 3,


How many terms of the G.P. $3, \frac{3}{2}, \frac{3}{4} \ldots . .$ are needed to give the sum $\frac{3069}{512} ?$


Here, $a=3$ and

Common ratio, $r=\frac{1}{2}$

And, $S_{n}=\frac{3069}{512}$

$\therefore S_{n}=3\left\{\frac{1-\left(\frac{1}{2}\right)^{n}}{1-\frac{1}{2}}\right\}$

$\Rightarrow \frac{3069}{512}=3\left\{\frac{1-\frac{1}{2^{n}}}{1-\frac{1}{2}}\right\}$

$\Rightarrow \frac{3069}{512}=6\left\{1-\frac{1}{2^{n}}\right\}$

$\Rightarrow \frac{3069}{3072}=1-\frac{1}{2^{n}}$

$\Rightarrow \frac{1}{2^{n}}=1-\frac{3069}{3072}$

$\Rightarrow \frac{1}{2^{n}}=\frac{3}{3072}$

$\Rightarrow \frac{1}{2^{n}}=\frac{3}{3072}$

$\Rightarrow 2^{n}=\frac{3072}{3}$

$\Rightarrow 2^{n}=1024$

$\Rightarrow 2^{n}=2^{10}$

$\therefore n=10$

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