Question

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

• A. $\frac{1}{3}\le {\alpha }^2\le 3$
• B. $\frac{1}{3}<{\alpha }^2<1$
• C. $1<\alpha <3$
• D. $\frac{1}{3}<\alpha <1$

Answer 1

Answer: (B)

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
$\therefore {\left({\alpha }^2+1\right)}^2-4{\left({\alpha }^2-1\right)}^2\ge 0$
$\left[{\alpha }^2+1+2\left({\alpha }^2-1\right)\right]\left[{\alpha }^2+1-2\left({\alpha }^2-1\right)\right]\ge 0$
$\Rightarrow \left(3{\alpha }^2-1\right)\left(3-{\alpha }^2\right)\ge 0$
$\Rightarrow 3{\alpha }^2-1\ge 0,3-{\alpha }^2\ge 0$
or $3{\alpha }^2-1\le 0,3-{\alpha }^2$
$\Rightarrow {\alpha }^2\ge \frac{1}{3},{\alpha }^2\le 3$
or ${\alpha }^2\le \frac{1}{3},{\alpha }^2\ge 3$ not possible
$\Rightarrow \frac{1}{3}\le {\alpha }^2\le 3$