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Question
How many terms of the G.P. 3, \[\frac{3}{2}, \frac{3}{4}\] ..... are needed to give the sum \[\frac{3069}{512}\] ?
Solution
\[\text { Here }, a = 3 \text { and }\]
\[\text { Common ratio }, r = \frac{1}{2} \]
\[\text { And, } S_n = \frac{3069}{512}\]
\[ \therefore S_n = 3\left\{ \frac{1 - \left( \frac{1}{2} \right)^n}{1 - \frac{1}{2}} \right\}\]
\[ \Rightarrow \frac{3069}{512} = 3\left\{ \frac{1 - \frac{1}{2^n}}{1 - \frac{1}{2}} \right\} \]
\[ \Rightarrow \frac{3069}{512} = 6 \left\{ 1 - \frac{1}{2^n} \right\}\]
\[ \Rightarrow \frac{3069}{3072} = 1 - \frac{1}{2^n} \]
\[ \Rightarrow \frac{1}{2^n} = 1 - \frac{3069}{3072} \]
\[ \Rightarrow \frac{1}{2^n} = \frac{3}{3072}\]
\[ \Rightarrow 2^n = \frac{3072}{3} \]
\[ \Rightarrow 2^n = 1024 \]
\[ \Rightarrow 2^n = 2^{10} \]
\[ \therefore n = 10\]
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