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(C) Let the first term be $a$ and the common ratio be $r$. The sum of an infinite $G.P.$ is given by
$S = \frac{a}{1-r}$, where $|r| < 1$.
The su

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$1/3$

Vedclass · Answer

(C) Let the first term be $a$ and the common ratio be $r$. The sum of an infinite $G.P.$ is given by
$S = \frac{a}{1-r}$, where $|r| < 1$.
The sum of the remaining terms after the first term is
$S - a = \frac{a}{1-r} - a$
$= a\left(\frac{1}{1-r} - 1ight)$
$= a\left(\frac{1 - (1-r)}{1-r}ight)$
$= a\left(\frac{r}{1-r}ight)$
$= \frac{ar}{1-r}$
According to the problem, the first term is equal to twice the sum of the remaining terms:
$a = 2\left(\frac{ar}{1-r}ight)$
Dividing both sides by $a$ (assuming $a 
eq 0$):
$1 = \frac{2r}{1-r}$
$1 - r = 2r$
$1 = 3r$
$r = \frac{1}{3}$

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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