Sequence And Series Question 465
Question: If $ \frac{1}{a},\frac{1}{b},\frac{1}{c} $ are A. P., then $ ( \frac{1}{a}+\frac{1}{b}-\frac{a}{c} ) $ $ ( \frac{1}{b}+\frac{1}{c}-\frac{1}{a} ) $ is equal to
Options:
A) $ \frac{4}{ac}-\frac{3}{b^{2}} $
B) $ \frac{b^{2}-ac}{a^{2}b^{2}c^{2}} $
C) $ \frac{4}{ac}-\frac{1}{b^{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \frac{1}{a}-\frac{1}{b}=\frac{1}{b}-\frac{1}{c} $
$ \therefore ( \frac{1}{a}+\frac{1}{b}-\frac{1}{c} )( \frac{1}{b}+\frac{1}{c}-\frac{1}{a} ) $ $ =( \frac{2}{a}-\frac{1}{b} )( \frac{2}{c}-\frac{1}{b} )=\frac{4}{ac}-\frac{1}{b}( \frac{2}{a}+\frac{2}{c} )+\frac{1}{b^{2}} $ $ =\frac{4}{ac}-\frac{2}{b}( \frac{2}{b} )+\frac{1}{b^{2}}=\frac{4}{ac}-\frac{3}{b^{2}} $
BETA
