Question

Suppose $x$ and $y$ are real numbers such that $-1 < x < y < 1$ Let $G$ be the sum of the geometric series whose first term is $x$ and whose common ratio is $y,$ and let $G^{\prime}$ be the sum of the geometric series whose first term is $y$ and common ratio is $x .$ If $G=G^{\prime},$ then the value of $(x+y)$ is

• A. 1
• B. -1
• C. 2
• D. -2

Answer 1

Answer: (A)

$G=\frac{x}{1-y}$ and $G'=\frac{y}{1-x} .$
As, $G=G' \Rightarrow \frac{x}{1-y}=\frac{y}{1-x}$
$x^{2}-x=y^{2}-y \Rightarrow (x+y-1)(x-y)=0$
So, we get $x+y-1=0$
$\Rightarrow x+v=1$