Question

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

Answer 1

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