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## Homework Statement

The sum of the squares of three distinct real numbers, which are in geometric progression is ##S^2##. If their sum is ##\alpha S##, show that ##\alpha^2 \in (1/3,1) \cup (1,3)##.

## Homework Equations

## The Attempt at a Solution

Let the three numbers be ##a , \, ar, \,ar^2##. According to the question

$$a^2+a^2r^2+a^2r^4=S^2 \, (*)$$

and

$$a+ar+ar^2=\alpha S \, (**)$$

Squaring (**), I get

$$a^2+a^2r^2+a^2r^4+2a^2r+2a^2r^3+2a^2r^2=\alpha^2S^2$$

From (*)

$$S^2+2a^2r(1+r+r^2)=\alpha^2S^2$$

The equation (**) can rewritten as ##a(1+r+r^2)=\alpha S##. Using this

$$S^2+2ar(\alpha S)=\alpha^2S^2 \Rightarrow S+2a\alpha r=\alpha^2S$$

I am stuck here.

Any help is appreciated. Thanks!