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Q.
The sum of the squares of three distinct real numbers which are in G.P. is $S^2$. If their sum is $\alpha \,S$ then
Sequences and Series
Solution:
Let $1, r, r^{2} $ be the three numbers in $G.P.$
$ \therefore 1+r+r^{2}= \alpha S$. Also $1+r^{2}+r^{4} = S^{2} $
$ \Rightarrow \left(1+r+r^{2}\right)\left(1-r+r^{2}\right) = S^{2} $
Also $\left(1+r+r^{2}\right)^{2}= \alpha^{2} S^{2}$
Dividing, we get $\frac{1-r+r^{2}}{1+r+r^{2}} = \frac{1}{\alpha^{2}}$
or $\alpha^{2} -\alpha^{2}r +\alpha^{2}r^{2} = 1+r+r^{2}$
$ r^{2} \left(\alpha^{2}-1\right) -\left(\alpha^{2}+1\right)r+\alpha^{2}-1= 0 $
Since $r$ is real and
$ \therefore \left(\alpha^{2}+1\right)^{2}-4\left(\alpha^{2}-1\right)^{2} \ge0 $
$\left[\alpha^{2}+1+2\left(\alpha^{2}-1\right)\right]\left[\alpha^{2}+1-2\left(\alpha^{2}-1\right)\right]\ge0$
$ \Rightarrow \left(3\alpha^{2}-1\right)\left(3-\alpha^{2}\right)\ge0$
$\Rightarrow 3\alpha^{2}-1\ge0, 3-\alpha^{2}\ge0 $
or $3\alpha^{2} -1\le0, 3-\alpha^{2}$
$ \Rightarrow \alpha^{2} \ge \frac{1}{3}, \alpha^{2}\le3$
or $\alpha^{2}\le\frac{1}{3}, \alpha^{2}\ge3$ not possible
$ \Rightarrow \frac{1}{3}\le \alpha^{2} \le3$