Question

If the A.M is twice the G.M. of the numbers $a$ and $b$, then $a:b$ will be

Answer 1

Find the Ratio and simplify

The two numbers are $a$ and $b$

$\mathrm{A}.\mathrm{M}=\frac{a+b}{2}$

$\mathrm{G}.\mathrm{M}=\sqrt{ab}$

The given condition is that A.M is twice the G.M.

$$\begin{gathered} \therefore \mathrm{A}.\mathrm{M}=2\mathrm{G}.\mathrm{M} \\ \Rightarrow \frac{\mathrm{a}+\mathrm{b}}{2}=2\sqrt{\mathrm{ab}} \\ \Rightarrow \frac{\mathrm{a}+\mathrm{b}}{2\sqrt{\mathrm{ab}}}=2 \\ \Rightarrow \frac{\mathrm{a}+\mathrm{b}+2\sqrt{\mathrm{ab}}}{\mathrm{a}+\mathrm{b}-2\sqrt{\mathrm{ab}}}=\frac{2+1}{2-1} \\ \Rightarrow \frac{{\left(\sqrt{a}+\sqrt{b}\right)}^2}{{\left(\sqrt{a}-\sqrt{b}\right)}^2}=\frac{3}{1}\left[{\left(a+b\right)}^2=a^2+2ab+b^2\right] \\ \Rightarrow \frac{\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)}=\frac{\sqrt{3}}{1} \\ \Rightarrow \frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{3}+1}{\sqrt{3}-1} \\ \Rightarrow \frac{a}{b}=\frac{{\left(\sqrt{3}+1\right)}^2}{{\left(\sqrt{3}-1\right)}^2} \\ \Rightarrow \frac{a}{b}=\frac{3+2\sqrt{3}+1}{3-2\sqrt{3}+1} \\ \Rightarrow \frac{a}{b}=\frac{4+2\sqrt{3}}{4-2\sqrt{3}} \\ \Rightarrow \frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}} \end{gathered}$$

Hence, the ratio $a:b$ is $2+\sqrt{3}:2-\sqrt{3}$