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 answers the following questions.If ${\alpha }^2=2$, then the value of $r$ equals-

• A. $\frac{1}{2}(5-\sqrt{3})$
• B. $\frac{1}{2}(3+\sqrt{5})$
• C. $\frac{1}{2}(\sqrt{5}+\sqrt{3})$
• D. $\frac{1}{3}(\sqrt{3}+\sqrt{5})$

Answer 1

The correct option is B $\frac{1}{2}(3+\sqrt{5})$

Let three number in $GP$ be $a,ar,{ar}^2.$

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

and $a+ar+{ar}^2=\alpha S$ ...(2)

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

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

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

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

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

As ${\alpha }^2=2\Rightarrow r=\frac{1}{2}(3+\sqrt{5})$