Question

The sum of the squares of three distinct real numbers, which are in GP, is $S^2$. If their sum is $aS$ then show that $a^2\in (\frac{1}{3},1)\cup (1,3)$

Answer 1

Let three numbers in G.P be $a,ar,a{r}^2$

$\Rightarrow a^2+a^2{r}^2+a^2{r}^4=S^2$ ......$\left(1\right)$

and $a+ar+a{r}^2=aS$ .........$\left(2\right)$

$\therefore \frac{a^2(1+r^2+r^4)}{a^2{(1+r+r^2)}^2}=\frac{S^2}{a^2{S}^2}$

$\Rightarrow \frac{{(1+r^2)}^2-r^2}{{(1+r+r^2)}^2}=\frac{1}{a^2}$

$\Rightarrow \frac{1+r^2-r}{1+r^2+r}=\frac{1}{a^2}$

$\Rightarrow a^2=\frac{r+\frac{1}{r}+1}{r+\frac{1}{r}-1}$

Put $r+\frac{1}{r}=y$

$\therefore \frac{y+1}{y-1}=a^2$

$\Rightarrow y+1=a^{2}y-a^2$

$\Rightarrow y=\frac{a^2+1}{a^2-1}$ since $\left|y\right|=|r+\frac{1}{r}|>2$

$\Rightarrow \left|\frac{a^2+1}{a^2-1}\right|>2$ where $a^2-1\ne 0$

$\Rightarrow |a^2+1|>2|a^2-1|$

$\Rightarrow {(a^2+1)}^2-{\left\{2(a^2-1)\right\}}^2>0$

$\Rightarrow \{(a^2+1)-2(a^2-1)\}\{(a^2+1)+2(a^2-1)\}>0$

$\Rightarrow (-a^2+3)(3a^2-1)>0$

$\therefore \frac{1}{3}<a^2<3$

$\therefore a^2\in (\frac{1}{3},1)\cup (1,3)$

$\because a^2\ne 1$