If a, b, and c are in H.P., then the value of $$(ac+ab$$−$$bc)(ab+bc$$−$$ac)(abc)2$$ is

$$1b+1c$$−$$1a1c+1a$$−$$1b=1b+1c$$−$$2b+1c1c+1b$$−$$1c=2c$$−$$1b1b=2bc$$−$$1b2Also$$ by eliminating b, we get the given $$expression(a+c)(3a$$−$$c)4a2c2$$

If a, b, and c are in H.P., then the value of $$(ac+ab$$−$$bc)(ab+bc$$−$$ac)(abc)2$$ is

1 b + 1 c − 1 a 1 c + 1 a − 1 b = 1 b + 1 c − 2 b + 1 c 1 c + 1 b − 1 c = 2 c − 1 b 1 b = 2 bc − 1 b 2 Also by eliminating b, we get the given expression ( a + c ) ( 3 a − c ) 4 a 2 c 2

Questions

If a, b, and c are in H.P., then the value of $\frac{(ac + ab - bc) (ab + bc - ac)}{(abc)^{2}}$ is

Unlock the full solution & master the concept.

Get a detailed solution and exclusive access to our masterclass to ensure you never miss a concept

By Expert Faculty of Sri Chaitanya

\frac{(a + c) (3 a - c)}{4 a^{2} c^{2}}

\frac{2}{bc} - \frac{1}{b^{2}}

\frac{2}{bc} - \frac{1}{a^{2}}

\frac{(a - c) (3 a + c)}{4 a^{2} c^{2}}

NEW

Ready to Test Your Skills?

Check Your Performance Today with our Free Mock Tests used by Toppers!

detailed solution

Correct option is A

$\begin{array}{l} (\frac{1}{b} + \frac{1}{c} - \frac{1}{a}) (\frac{1}{c} + \frac{1}{a} - \frac{1}{b}) = (\frac{1}{b} + \frac{1}{c} - \frac{2}{b} + \frac{1}{c}) (\frac{1}{c} + \frac{1}{b} - \frac{1}{c}) \\ = (\frac{2}{c} - \frac{1}{b}) \frac{1}{b} = \frac{2}{bc} - \frac{1}{b^{2}} \end{array}$
Also by eliminating b, we get the given expression
$\frac{(a + c) (3 a - c)}{4 a^{2} c^{2}}$