Sequence And Series Question 14
Question: If the harmonic mean between $ a $ and $ b $ be $ H $ , then the value of $ \frac{1}{H-a}+\frac{1}{H-b} $ is
Options:
A) $ a+b $
B) $ ab $
C) $ \frac{1}{a}+\frac{1}{b} $
D) $ \frac{1}{a}-\frac{1}{b} $
Show Answer
Answer:
Correct Answer: C
Solution:
Putting $ H=\frac{2ab}{a+b} $ , we have $ \frac{1}{H-a}+\frac{1}{H-b} $ $ =\frac{1}{( \frac{2ab}{a+b}-a )}+\frac{1}{( \frac{2ab}{a+b}-b )}=\frac{a+b}{ab-a^{2}}+\frac{a+b}{ab-b^{2}} $ $ =( \frac{a+b}{b-a} )( \frac{1}{a}-\frac{1}{b} )=( \frac{a+b}{b-a} )( \frac{b-a}{ab} )=\frac{a+b}{ab}=\frac{1}{a}+\frac{1}{b} $ .
BETA
