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Question
The 4th and 7th terms of a G.P. are \[\frac{1}{27} \text { and } \frac{1}{729}\] respectively. Find the sum of n terms of the G.P.
Solution
Let a be the first term and r be the common ratio of the G.P.
\[\therefore a_4 = \frac{1}{27} \]
\[ \Rightarrow a r^{4 - 1} = \frac{1}{27}\]
\[ \Rightarrow a r^3 = \frac{1}{27} \]
\[ \Rightarrow \left( a r^3 \right)^2 = \frac{1}{{27}^2}\]
\[ \Rightarrow a^2 r^6 = \frac{1}{729} \]
\[ \Rightarrow a r^6 = \frac{1}{729a} . . . \left( i \right)\]
\[\text {Similarly }, a_7 = \frac{1}{729} \]
\[ \Rightarrow a r^{7 - 1} = \frac{1}{729}\]
\[ \Rightarrow a r^6 = \frac{1}{729} \]
\[ \Rightarrow a r^6 = \frac{1}{729a} \left[ \text { From } \left( i \right) \right] \]
\[ \therefore a = 1\]
\[\text { Putting this in } a_4 = \frac{1}{27}\]
\[ \Rightarrow a r^3 = \frac{1}{3^3}\]
\[ \Rightarrow r^3 = \frac{1}{3^3} \]
\[ \therefore r = \frac{1}{3}\]
\[\text { Now, sum of n terms of the G . P } . , S_n = a\left( \frac{r^n - 1}{r - 1} \right)\]
\[ \Rightarrow S_n = 1\left( \frac{1 - \left( \frac{1}{3} \right)^n}{1 - \frac{1}{3}} \right) \]
\[ \Rightarrow S_n = \frac{3}{2}\left( 1 - \frac{1}{3^n} \right)\]
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