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Question

Mathematics Question on Arithmetic Progression

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

A

4ac−3b2\frac{4}{ac} - \frac{3}{b^{2}}

B

b2−aca2b2c2\frac{b^{2}-ac}{a^{2}b^{2}c^{2}}

C

4ac=1b2\frac{4}{ac} = \frac{1}{b^{2}}

D

None of these

Answer

4ac−3b2\frac{4}{ac} - \frac{3}{b^{2}}

Explanation

Solution

The correct answer is A:4ac−3b2\frac{4}{ac}-\frac{3}{b^2}
Given that:
1a,1b  and  1c\frac{1}{a},\frac{1}{b}\space and\space \frac{1}{c} are in A.P,
we know that, if the series is in A.P, then the difference of the consecutive terms are equal
i.e, 1b−1a=1c−1b\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b} -(i)
∴\therefore as per the question we need to find:
(1a+1b−1c)(1b+1c−1a)(\frac{1}{a}+\frac{1}{b}-\frac{1}{c})(\frac{1}{b}+\frac{1}{c}-\frac{1}{a})
=(1a+1b−1a)(1c+1c−16)=(\frac{1}{a}+\frac{1}{b}-\frac{1}{a})(\frac{1}{c}+\frac{1}{c}-\frac{1}{6}) [from equation (i)]
=(2a−1b)(2c−1b)=(\frac{2}{a}-\frac{1}{b})(\frac{2}{c}-\frac{1}{b})
=4ac−2ab−2bc+1b2=\frac{4}{ac}-\frac{2}{ab}-\frac{2}{bc}+\frac{1}{b^2}
=4ac−2b(1a+1c)+1b2=\frac{4}{ac}-\frac{2}{b}(\frac{1}{a}+\frac{1}{c})+\frac{1}{b^2}
=4ac−4b2+1b2=\frac{4}{ac}-\frac{4}{b^2}+\frac{1}{b^2} (∴1a+1c=2b)(\therefore \frac{1}{a}+\frac{1}{c}=\frac{2}{b})
=4ac−3b2=\frac{4}{ac}-\frac{3}{b^2}