If 1-a- 1-b- 1-c are in AP- then 1-a-1-b-1-c1-b-1-c-1-a is equal toA- 4-a c-3-b2B- b2-a c-a2 b2 c2C- 4-a c-1-b2D- None of these

The correct option is A ( 1/a+1/b+1/c 2/c ) ( 1/a+1/b+1/c 2/a ) = ( 3/b 2/c ) ( 3/b 2/a ) = 9/b^2 6/ab 6/bc+4/ac = 9/b^2 6/ab ( 1/a+1/c )+ 4/ac = 4 ...

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Byju's Answer

Standard XII

Mathematics

Sum of Infinite Terms of a GP

If 1/a, 1/b, ...

Question

If $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in AP, then $(\frac{1}{a} + \frac{1}{b} - \frac{1}{c}) (\frac{1}{b} + \frac{1}{c} - \frac{1}{a})$ is equal to

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Solution

The correct option is A
$(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - \frac{2}{c}) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - \frac{2}{a}) = (\frac{3}{b} - \frac{2}{c}) (\frac{3}{b} - \frac{2}{a}) = \frac{9}{b^{2}} - \frac{6}{a b} - \frac{6}{b c} + \frac{4}{a c} = \frac{9}{b^{2}} - \frac{6}{a b} (\frac{1}{a} + \frac{1}{c}) + \frac{4}{a c} = \frac{4}{a c} - \frac{3}{b^{2}}$

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If 1a,1b,1c are in AP, then (1a+1b−1c)(1b+1c−1a) is equal to

The correct option is A (1a+1b+1c−2c)(1a+1b+1c−2a)=(3b−2c)(3b−2a)=9b2−6ab−6bc+4ac=9b2−6ab(1a+1c)+4ac=4ac−3b2

If $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in AP, then $(\frac{1}{a} + \frac{1}{b} - \frac{1}{c}) (\frac{1}{b} + \frac{1}{c} - \frac{1}{a})$ is equal to