Question

If AM of two numbers is twice of their GM, then the ratio of greatest number to smallest number is

• A. $7-4\sqrt{3}$
• B. $7+4\sqrt{3}$
• C. $21$
• D. $5$

Answer 1

The correct option is B

$7+4\sqrt{3}$

Explanation for the correct option:

Step 1. Let $a$ and $b$ be the two numbers.

Given, AM $=2$ GM

$\Rightarrow$$\frac{(a+b)}{2}=2\surd \left(ab\right)$

$\Rightarrow$ $(a+b)=4\surd \left(ab\right)$

Step 2. By Squaring both sides, we get

${(a+b)}^2=16ab$ ....(i)

$\Rightarrow$$a^2+b^2-14ab=0$

$\Rightarrow$ $a^2+b^2-2ab=12ab$

$\Rightarrow$ ${(a-b)}^2=12ab$ ....(ii)

Step 3. By Dividing equation (i) by (ii), we get

$\frac{{(a+b)}^2}{{(a-b)}^2}=\frac{16ab}{12ab}$

$\Rightarrow$$\frac{{(a+b)}^2}{{(a-b)}^2}=\frac{4}{3}$

Take square root on both sides,

$\frac{(a+b)}{(a-b)}=\frac{2}{\surd 3}$

Step 4. By Using componendo dividendo rule, we get

$\frac{(a+b+a-b)}{(a+b-a+b)}=\frac{(2+\surd 3)}{(2-\surd 3)}$

$\Rightarrow$ $\frac{2a}{2b}=\frac{(2+\surd 3)}{(2-\surd 3)}$

$\Rightarrow$ $\frac{a}{b}=\frac{(2+\surd 3)}{(2-\surd 3)}$

$\Rightarrow$ $\frac{a}{b}=\frac{(2+\surd 3)(2+\surd 3)}{(2-\surd 3)(2+\surd 3)}$

$$\begin{gathered} =\frac{{(2+\surd 3)}^2}{(4-3)} \\ ={(2+\surd 3)}^2=4+4\surd 3+3 \\ =\mathbf{7}\mathbf{+}\mathbf{4}\mathbf{\surd }\mathbf{3} \end{gathered}$$

Hence, Option ‘B’ is Correct.