Sequence And Series Question 231
Question: If the A.M. is twice the G.M. of the numbers $ a $ and $ b $ , then $ a:b $ will be
[Roorkee 1953]
Options:
A) $ \frac{2-\sqrt{3}}{2+\sqrt{3}} $
B) $ \frac{2+\sqrt{3}}{2-\sqrt{3}} $
C) $ \frac{\sqrt{3}-2}{\sqrt{3}+2} $
D) $ \frac{\sqrt{3}+2}{\sqrt{3}-2} $
Show Answer
Answer:
Correct Answer: B
Solution:
Given A.M. $ = $ 2(G.M.) or $ \frac{1}{2}(a+b)=2\sqrt{ab} $ or $ \frac{a+b}{2\sqrt{ab}}=\frac{2}{1} $
$ \Rightarrow $ $ \frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\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}={{( \frac{\sqrt{3}+1}{\sqrt{3}-1} )}^{2}} $
$ \Rightarrow $ $ \frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}} $ or $ a:b=(2+\sqrt{3}):(2-\sqrt{3}) $ .
BETA
