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The 4th and 7th Terms of a G.P. Are 1 27 and 1 729 Respectively. Find the Sum of N Terms of the G.P. - Mathematics

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.

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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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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 20 Geometric Progression
Exercise 20.3 | Q 11 | Page 28
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