Suppose x and y are real numbers such that -1

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

Tardigrade Admin · Accepted Answer

G=(x/1-y) and G'=(y/1-x) . As, G=G' ⇒ (x/1-y)=(y/1-x) x2-x=y2-y ⇒ (x+y-1)(x-y)=0 So, we get x+y-1=0 ⇒ x+v=1

Q. 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′ be the sum of the geometric series whose first term is y and common ratio is x. If G=G′, then the value of (x+y) is

1

-1

2

-2

G=1−yx​ and G′=1−xy​. As, G=G′⇒1−yx​=1−xy​ x2−x=y2−y⇒(x+y−1)(x−y)=0 So, we get x+y−1=0 ⇒x+v=1

$G = \frac{x}{1 - y}$ 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 $x + y - 1 = 0$
$\Rightarrow x + v = 1$