The sum of squares of 3 distinct real numbers which are in G.P. is $s^{2}$. If their sum is $\alpha$

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The sum of squares of 3 distinct real numbers which are in G.P. is $s^{2}$. If their sum is $\alpha$ s, then $\alpha^{2}$ lie in interval is

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Sol. [Key Concept: Three in G.P. with fixed sum-of-squares and sum ⇒ constrain the common ratio / interval.]$$\alpha s = a\left( 1 + r + r^{2} ight)$$$$\begin{matrix} & s^{2} = a^{2}\left( 1 + r^{2} + r^{4} ight) \ \Rightarrow & \alpha^{2} = \frac{\left( 1 + r + r^{2} ight)}{1 - r + r^{2}} = 1 + \frac{2r}{1 - r + r^{2}} = 1 + \frac{2}{\frac{1}{r} + r - 1} \ \Rightarrow & \alpha^{2} \in \left( 1 - \frac{2}{3},1 - 0 ight) \cup (1 + 0,1 + 2) \ \Rightarrow & \alpha^{2} \in \left( \frac{1}{3},1 ight) \cup (1,3) \end{matrix}$$

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