Sequence And Series Question 360
Question: If a, b and c are in H. P then the value of $ ( \frac{1}{b}+\frac{1}{c}-\frac{1}{a} )( \frac{1}{c}+\frac{1}{a}-\frac{1}{b} ) $ is:
Options:
A) $ \frac{2}{bc}+\frac{1}{b^{2}} $
B) $ \frac{3}{c^{2}}+\frac{2}{ca} $
C) $ \frac{3}{b^{2}}-\frac{2}{ab} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Let a, b and c are H.P.
$ \therefore ,\frac{1}{a},\frac{1}{b},\frac{1}{c} $ are in A.P.
$ \therefore ,\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b} $
$ \Rightarrow \frac{1}{c}=\frac{2}{b}-\frac{1}{a} $ Consider $ ( \frac{1}{b}+\frac{1}{c}-\frac{1}{a} )( \frac{1}{c}+\frac{1}{a}-\frac{1}{b} ) $ $ =( \frac{1}{b}+\frac{2}{b}-\frac{1}{a}-\frac{1}{a} )( \frac{2}{b}-\frac{1}{b} ) $ Using $ \frac{1}{a}+\frac{1}{c}=\frac{2}{b}=( \frac{3}{b}-\frac{2}{a} )( \frac{1}{b} )=\frac{3}{b^{2}}-\frac{2}{ab} $
BETA
