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How Many Terms of the G.P. 3, 3/2, 3/4, ... Be Taken Together to Make 3069 512 ? - Mathematics

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प्रश्न

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

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उत्तर

Here, a = 3
Common ratio, 

\[r = \frac{1}{2}\]

Sn = \[\frac{3069}{512}\]

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

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

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

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

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

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

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

\[ \Rightarrow 2^n = 1024 \]

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

\[ \therefore n = 10\]

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अध्याय 20: Geometric Progression - Exercise 20.3 [पृष्ठ २८]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 20 Geometric Progression
Exercise 20.3 | Q 5 | पृष्ठ २८

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