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 ${\alpha }^2$ lies in,

• A. $(\frac{1}{3},2)$
• B. $(1,2)$
• C. $(\frac{1}{3},3)$
• D. none of these

Answer 1

Answer: (C)

$(\frac{1}{3},3)$

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

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

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

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

or $r^2-\left(\frac{{\alpha }^2+1}{{\alpha }^2-1}\right)r+1=0$

$({\alpha }^2\ne 1)$

$\because r$ is real,${\left(\frac{{\alpha }^2+1}{{\alpha }^2-1}\right)}^2-\mathrm{4.1.1}>0$

$\because$(The numbers in G.P are distinct)

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

$\Rightarrow (3{\alpha }^2-1)(-{\alpha }^2+3)>0$

or $({\alpha }^2-\frac{1}{3})({\alpha }^2-3)<0$

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

But ${\alpha }^2\ne 1$

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

Hence option $C$ is the answer.