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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2014 · Mathematics · Sequences and Series

If 1 a , 1 ~b , 1 c are in A. P., then ( 1 a + 1 ~b - 1 c )...

If 1a,1 b,1c\frac{1}{\mathrm{a}}, \frac{1}{\mathrm{~b}}, \frac{1}{\mathrm{c}} are in A. P., then \left(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}-\frac{1}{\mathrm{c}}\right) $\left(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\right) is equal to

Options

  1. A4ac−3b2\frac{4}{a c}-\frac{3}{b^{2}}
  2. Bb2−aca2b2c2\frac{b^{2}-a c}{a^{2} b^{2} c^{2}}
  3. C4ac−1b2\frac{4}{a c}-\frac{1}{b^{2}}
  4. DNone of these

Correct answer

A. 4ac−3b2\frac{4}{a c}-\frac{3}{b^{2}}

Step-by-step solution

\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{~b}}=\frac{1}{\mathrm{~b}}-\frac{1}{\mathrm{c}} \therefore\left(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}-\frac{1}{\mathrm{c}}\right)\left(\frac{1}{\mathrm{~b}}+\frac{1}{\mathrm{c}}-\frac{1}{\mathrm{a}}\right)=(2a−1 b)(2c−1 b)=4ac−1 b(2a+2c)+1 b2=\left(\frac{2}{\mathrm{a}}-\frac{1}{\mathrm{~b}}\right)\left(\frac{2}{\mathrm{c}}-\frac{1}{\mathrm{~b}}\right)=\frac{4}{\mathrm{ac}}-\frac{1}{\mathrm{~b}}\left(\frac{2}{\mathrm{a}}+\frac{2}{\mathrm{c}}\right)+\frac{1}{\mathrm{~b}^{2}} =4ac−2 b(2 b)+1 b2=4ac−3 b2=\frac{4}{\mathrm{ac}}-\frac{2}{\mathrm{~b}}\left(\frac{2}{\mathrm{~b}}\right)+\frac{1}{\mathrm{~b}^{2}}=\frac{4}{\mathrm{ac}}-\frac{3}{\mathrm{~b}^{2}}
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