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 $S = \frac{a}{1 - r} = 9$.From this,we get $a =

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$2/3$

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 $S = \frac{a}{1 - r} = 9$.From this,we get $a = 9(1 - r)$ $\dots(i)$.The sum of the first two terms is $a + ar = 5$,which can be written as $a(1 + r) = 5$ $\dots(ii)$.Substituting the value of $a$ from equation $(i)$ into equation $(ii)$:$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 options provided include $2/3$,the correct option is $D$.

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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