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How many terms of the G.P. 3, 3/2, 3/4 ..... are needed to give the sum 3069/512?

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प्रश्न

How many terms of the G.P. `3, 3/2, 3/4` ..... are needed to give the sum `3069/512`?

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उत्तर

\[\text { Here }, a = 3 \text { and }\]

\[\text { Common ratio }, r = \frac{1}{2} \]

\[\text { And, } S_n = \frac{3069}{512}\]

\[ \therefore S_n = 3\left\{ \frac{1 - \left( \frac{1}{2} \right)^n}{1 - \frac{1}{2}} \right\}\]

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

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

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

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

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

\[ \Rightarrow 2^n = \frac{3072}{3} \]

\[ \Rightarrow 2^n = 1024 \]

\[ \Rightarrow 2^n = 2^{10} \]

\[ \therefore n = 10\]

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अध्याय 9: Arithmetic and geometric progression - CHAPTER TEST [पृष्ठ २०२]

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नूतन Mathematics [English] Class 10 ICSE
अध्याय 9 Arithmetic and geometric progression
CHAPTER TEST | Q 8. | पृष्ठ २०२

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