Question

The sum of three terms of a strictly increasing G.P is $\alpha S$ and the sum of the squares of these terms is $S^2$, then if we drop the condition, that the Geometric progression is strictly increasing and take $r^2=1$, (where $r$ is the common ratio of G.P) then the value of $\alpha$ is,

• A. $0$
• B. $\pm 1$
• C. $\pm 2$
• D. $\pm \sqrt{3}$

Answer 1

The correct option is D $\pm \sqrt{3}$

Let the three numbers in strictly increasing G.P are $\frac{a}{r},a,ar(r>1)$

According to the passage $\frac{a^2}{r^2}+a^2+a^2{r}^2=S^2$

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

Splitting as $\Rightarrow a^2(\frac{1}{r}+1+r)(\frac{1}{r}-1+r)=S^2$ ....$\left(1\right)$

and $\frac{a}{r}+a+ar=\alpha S$

Squaring both sides, we get$a^2{(\frac{1}{r}+1+r)}^2={\alpha }^2{S}^2$ .......$\left(2\right)$

Dividing eqn$\left(2\right)$ by $\left(1\right)$, then

$\left(\frac{\frac{1}{r}+1+r}{\frac{1}{r}-1+r}\right)={\alpha }^2$

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

$\Rightarrow ({\alpha }^2-1)r^2-({\alpha }^2+1)r+{\alpha }^2-1=0$ .......$\left(3\right)$

Put $r^2=1$ or $r=\pm 1$ in the eqn$\left(3\right)$ then

$({\alpha }^2-1)\pm ({\alpha }^2+1)+{\alpha }^2-1=0$ ....$\left(4\right)$

Taking the positive sign as it is an increasing A.P

$\Rightarrow 3{\alpha }^2-1=0$

$\therefore {\alpha }^2=\frac{1}{3}$

or $\alpha =\pm \frac{1}{\sqrt{3}}$

and taking the negative sign from $\left(4\right)$, we get ${\alpha }^2=3$

or $\alpha =\pm \sqrt{3}$

Hence option $D$ is the answer