Question

How many terms of the G.P. 3, 3/2, 3/4, ... be taken together to make $\frac{3069}{512}$?

Answer 1

Here, a = 3
Common ratio, $r=\frac{1}{2}$
S_n = $\frac{3069}{512}$
$$\begin{gathered} \therefore S_n=3\left\{\frac{1-{\left({\displaystyle \frac{1}{2}}\right)}^n}{1-{\displaystyle \frac{1}{2}}}\right\} \\ \Rightarrow \frac{3069}{512}=3\left\{\frac{1-{\displaystyle \frac{1}{2^n}}}{{\displaystyle \frac{1}{2}}}\right\} \\ \Rightarrow \frac{3069}{512}=6\left\{1-\frac{1}{2^n}\right\} \\ \Rightarrow \frac{3069}{3072}=1-\frac{1}{2^n} \\ \Rightarrow \frac{1}{2^n}=1-\frac{3069}{3072} \\ \Rightarrow \frac{1}{2^n}=\frac{3}{3072} \\ \Rightarrow 2^n=\frac{3072}{3} \\ \Rightarrow 2^n=1024 \\ \Rightarrow 2^n=2^{10} \\ \therefore n\mathit{}=10 \end{gathered}$$