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(C) Let the infinite $G.P.$ be $a, ar, ar^2, ar^3, \dots$ where $|r| < 1$.The sum of the remaining terms starting from the second term is given by $S_

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$1/3$

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(C) Let the infinite $G.P.$ be $a, ar, ar^2, ar^3, \dots$ where $|r| The sum of the remaining terms starting from the second term is given by $S_{	ext{remaining}} = ar + ar^2 + ar^3 + \dots = \frac{ar}{1-r}$.According to the problem,the first term $a$ is equal to twice the sum of the remaining terms:$a = 2 \left( \frac{ar}{1-r} ight)$.Assuming $a 
eq 0$,we can divide both sides by $a$:$1 = \frac{2r}{1-r}$.Multiplying both sides by $(1-r)$:$1 - r = 2r$.Rearranging the terms to solve for $r$:$1 = 3r \Rightarrow r = \frac{1}{3}$.

If in an infinite $G.P.$ the first term is equal to twice the sum of the remaining terms,then its common ratio is

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