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 $7+4\sqrt{3}$

Let the two numbers be $a\&b$.

Then,

According to given question,

$A.M.=\frac{a+b}{2}\text{and}\text{G}\text{.M}\text{.=}\sqrt{ab}$

Given that,

$A.M.=2\times G.M.$

Now,

$\frac{a+b}{2}=2\sqrt{ab}$

$a+b=4\sqrt{ab}$

Squaring both side and we get,

${(a+b)}^2=16ab......\left(1\right)$

$a^2+b^2+2ab=16ab$

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

$a^2+b^2=-2ab-12ab=0$

${(a-b)}^2=12ab......\left(2\right)$

Divided equation (1) by (2) and we get,

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

$\frac{a+b}{a-b}=\frac{2}{\sqrt{3}}$

Using componento and dividendo rule……

$\frac{a+b+a-b}{a+b-a+b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$

$\frac{2a}{2b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$

$\frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$

$\frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}$

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

$\frac{a}{b}=4+3+4\sqrt{3}$

$\frac{a}{b}=7+4\sqrt{3}$

Hence, this is the answer.