The sum of an infinite number of terms in a G | vedclass

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(B) Let the first term of the $G.P.$ be $a$ and the common ratio be $r$. Given that the sum of an infinite $G.P.$ is $S = \frac{a}{1-r} = 20$ .....$(i

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$3/5$

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(B) Let the first term of the $G.P.$ be $a$ and the common ratio be $r$. Given that the sum of an infinite $G.P.$ is $S = \frac{a}{1-r} = 20$ .....$(i)$ The squares of the terms form a new $G.P.$ with first term $a^2$ and common ratio $r^2$. The sum of the squares is $S' = \frac{a^2}{1-r^2} = 100$ .....$(ii)$ From $(i)$,we have $a = 20(1-r)$. Substitute this into $(ii)$: $\frac{[20(1-r)]^2}{1-r^2} = 100$ $\frac{400(1-r)^2}{(1-r)(1+r)} = 100$ $\frac{4(1-r)}{1+r} = 1$ $4 - 4r = 1 + r$ $5r = 3$ $r = 3/5$.

The sum of an infinite number of terms in a $G.P.$ is $20$ and the sum of their squares is $100$. The common ratio of the $G.P.$ is:

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