Sequence And Series Question 256
Question: If the ratio of A.M. between two positive real numbers $ a $ and $ b $ to their H.M. is $ m:n $ , then $ a:b $ is
Options:
A) $ \frac{\sqrt{m-n}+\sqrt{n}}{\sqrt{m-n}-\sqrt{n}} $
B) $ \frac{\sqrt{n}+\sqrt{m-n}}{\sqrt{n}-\sqrt{m-n}} $
C) $ \frac{\sqrt{m}+\sqrt{m-n}}{\sqrt{m}-\sqrt{m-n}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{\frac{a+b}{2}}{\frac{2ab}{a+b}}=\frac{m}{n} $
$ \Rightarrow $ $ \frac{{{(a+b)}^{2}}}{4ab}=\frac{m}{n} $
$ \Rightarrow $ $ \frac{{{(a+b)}^{2}}}{2ab}=\frac{2m}{n} $ Applying dividendo, we get $ \frac{a^{2}+b^{2}}{2ab}=\frac{2m-n}{n} $ Applying componendo and dividendo, We get $ \frac{{{(a+b)}^{2}}}{{{(a-b)}^{2}}}=\frac{m}{m-n} $
$ \Rightarrow $ $ \frac{a+b}{a-b}=\frac{\sqrt{m}}{\sqrt{m-n}} $ Again, applying componendo and dividendo We get $ \frac{a}{b}=\frac{\sqrt{m}+\sqrt{m-n}}{\sqrt{m}-\sqrt{m-n}} $ .
BETA
