If the sum of an infinite GP is 9 and the sum | vedclass

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Question

(C) Let the first term be $a$ and the common ratio be $r$. Given that the sum of an infinite $GP$ is $S_{\infty} = \frac{a}{1-r} = 9$. This implies $a

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(C) Let the first term be $a$ and the common ratio be $r$. Given that the sum of an infinite $GP$ is $S_{\infty} = \frac{a}{1-r} = 9$. This implies $a = 9(1-r)$. The sum of the first two terms is $a + ar = 5$. Substituting $a = 9(1-r)$ into the equation: $9(1-r)(1+r) = 5$. $9(1-r^2) = 5$. $1-r^2 = \frac{5}{9}$. $r^2 = 1 - \frac{5}{9} = \frac{4}{9}$. Thus,$r = \pm \frac{2}{3}$. Since the options provided include $\frac{2}{3}$,the correct common ratio is $\frac{2}{3}$.

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

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