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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 .....

Similar Questions

The roots of the equation $x^3-14x^2+56x-64=0$ are in

If ${x_1}, {x_2}, {x_3}$ as well as ${y_1}, {y_2}, {y_3}$ are in $G$.$P$. with the same common ratio,then the points $({x_1}, {y_1}), ({x_2}, {y_2})$ and $({x_3}, {y_3})$:

The two geometric means between $1$ and $64$ are ........

The sum of the $3^{rd}$ and the $4^{th}$ terms of a $G.P.$ is $60$ and the product of its first three terms is $1000$. If the first term of this $G.P.$ is positive,then its $7^{th}$ term is

The $6^{th}$ term of a $G.P.$ is $32$ and its $8^{th}$ term is $128$,then the common ratio of the $G.P.$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module