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Question
Mark the correct alternative in the following question:
Let S be the sum, P be the product and R be the sum of the reciprocals of 3 terms of a G.P. Then p2R3 : S3 is equal to
Options
(a) 1 : 1
(b) (Common ratio)n : 1
(c) (First term)2 : (Common ratio)2
(d) None of these
Solution
\[\text{ Let the three terms of the G . P . be \frac{a}{r}, a, ar . Then }\]
\[S = \frac{a}{r} + a + ar\]
\[ = a\left( \frac{1}{r} + 1 + r \right)\]
\[ = a\left( \frac{1 + r + r^2}{r} \right)\]
\[ = \frac{a\left( r^2 + r + 1 \right)}{r}\]
\[\text{ Also }, \]
\[P = \frac{a}{r} \times a \times ar = a^3 \]
\[\text{ And }, \]
\[R = \frac{r}{a} + \frac{1}{a} + \frac{1}{ar}\]
\[ = \frac{1}{a}\left( r + 1 + \frac{1}{r} \right)\]
\[ = \frac{1}{a}\left( \frac{r^2 + r + 1}{r} \right)\]
\[\text{ Now }, \]
\[\frac{P^2 R^3}{S^3} = \frac{\left( a^3 \right)^2 \times \left[ \frac{1}{a}\left( \frac{r^2 + r + 1}{r} \right) \right]^3}{\left[ a\left( \frac{r^2 + r + 1}{r} \right) \right]^3}\]
\[ = \frac{a^6 \times \frac{1}{a^3} \left( \frac{r^2 + r + 1}{r} \right)^3}{a^3 \left( \frac{r^2 + r + 1}{r} \right)^3}\]
\[ = \frac{1}{1}\]
\[\text{ So, the ratio is }1: 1 .\]
Hence, the correct alternative is option (a).
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