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

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

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$3/5$

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(B) Let the first term be $a$ and the common ratio be $r$. The sum of an infinite $G.P.$ is given by $S = \frac{a}{1-r} = 20 \quad (i)$.The sum of the squares of the terms is given by $\frac{a^2}{1-r^2} = 100 \quad (ii)$.We can rewrite $(ii)$ as $\frac{a}{1-r} 	imes \frac{a}{1+r} = 100$.Substituting $(i)$ into this equation: $20 	imes \frac{a}{1+r} = 100$,which simplifies to $\frac{a}{1+r} = 5 \quad (iii)$.From $(i)$,$a = 20(1-r)$. Substituting this into $(iii)$:$\frac{20(1-r)}{1+r} = 5$$4(1-r) = 1+r$$4 - 4r = 1 + r$$3 = 5r$$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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