Search any question & find its solution
Question:
Answered & Verified by Expert
If the A.M. is twice the G.M. of the numbers $a$ and $b$, then $a: b$ will be
Options:
Solution:
1377 Upvotes
Verified Answer
The correct answer is:
$\frac{2+\sqrt{3}}{2-\sqrt{3}}$
Given A.M. $=2$ (G.M.) or $\frac{1}{2}(a+b)=2 \sqrt{a b}$
$\frac{a+b}{2 \sqrt{a b}}=\frac{2}{1}$ $\Rightarrow \frac{a+b+2 \sqrt{a b}}{a+b-2 \sqrt{a b}}$ $=\frac{2+1}{2-1}=\frac{3}{1}$
$\Rightarrow \frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}=\frac{3}{1} \Rightarrow$ $\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{3}}{1}$
$\Rightarrow \frac{a}{b}=\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)^2$
$\Rightarrow \quad \frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$ $a: b=(2+\sqrt{3}):(2-\sqrt{3})$
$\frac{a+b}{2 \sqrt{a b}}=\frac{2}{1}$ $\Rightarrow \frac{a+b+2 \sqrt{a b}}{a+b-2 \sqrt{a b}}$ $=\frac{2+1}{2-1}=\frac{3}{1}$
$\Rightarrow \frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}=\frac{3}{1} \Rightarrow$ $\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{3}}{1}$
$\Rightarrow \frac{a}{b}=\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)^2$
$\Rightarrow \quad \frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$ $a: b=(2+\sqrt{3}):(2-\sqrt{3})$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.