If the sum of an infinite G.P. is 9 and the s | vedclass

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(D) Let the infinite $G.P.$ be $a, ar, ar^2, \dots, \infty$.The sum of an infinite $G.P.$ is given by $S = \frac{a}{1-r} = 9$,where $|r| < 1$.$\Righta

Vedclass · Accepted Answer

$2/3$

Vedclass · Answer

(D) Let the infinite $G.P.$ be $a, ar, ar^2, \dots, \infty$.The sum of an infinite $G.P.$ is given by $S = \frac{a}{1-r} = 9$,where $|r| $\Rightarrow a = 9(1-r) \dots (i)$The sum of the first two terms is $a + ar = 5$.$\Rightarrow a(1+r) = 5 \dots (ii)$Substituting $(i)$ into $(ii)$:$9(1-r)(1+r) = 5$$9(1-r^2) = 5$$1-r^2 = \frac{5}{9}$$r^2 = 1 - \frac{5}{9} = \frac{4}{9}$$r = \pm \frac{2}{3}$.Since the sum of the infinite series exists,$|r| < 1$,both values are valid. Given the options,the correct choice is $2/3$.

If the sum of an infinite $G.P.$ is $9$ and the sum of the first two terms is $5$,then the common ratio is

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