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 ${\alpha }^2=2$, then the value of $\left[r\right]$ is (where$[.]$ denotes the Greatest Integer function and $r$ is the common ratio of the Geometric progression.)

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (C)

$2$

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 ${\alpha }^2=2$ in eqn$\left(3\right)$, then$r^2-3r+1=0$

Using the formula for root of a quadratic equation

$=\frac{-b\pm \sqrt{b^2-4ac}}{2}$

$\therefore r=\frac{-3\pm \sqrt{3^2-4\times 1\times 1}}{2}=\frac{3\pm \sqrt{5}}{2}=\frac{3+\sqrt{5}}{2}=\mathrm{2.61..}$

$\therefore$ GIF$\left[r\right]=2$

Hence option $C$ is the answer