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 ratio is y, and let G′ be the sum of the geometric series whose first term is y and ratio is x. If G=G′, find x+y.
A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.
Solution
We note that G=x/(1−y) and G′=y/(1−x). Setting them equal gives x/(1−y)=y/(1−x)⇒x2−x=y2−x⇒(x+y−1)(x−y)=0, so we get that x+y−1=0⇒x+y=1.
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