Question

If $a,b,c$ are in H.P,then the expression $E=(\frac{1}{b}+\frac{1}{c}-\frac{1}{a})(\frac{1}{c}+\frac{1}{a}-\frac{1}{b})$ equals

• A. $\frac{2}{bc}-\frac{1}{b^2}$
• B. $\frac{1}{4}(\frac{3}{c^2}+\frac{2}{ca}-\frac{1}{a^2})$
• C. $\frac{3}{b^2}-\frac{2}{ab}$
• D. $Noneofthese.$

Answer 1

Answer: (A), (B), (C)

$\frac{2}{bc}-\frac{1}{b^2}$
$\frac{1}{4}(\frac{3}{c^2}+\frac{2}{ca}-\frac{1}{a^2})$
$\frac{3}{b^2}-\frac{2}{ab}$

As $a,b,c$ are in H.P., $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.
Let the common difference of this A.P. be d Now,

$E=(\frac{1}{c}+d)(\frac{1}{c}-d)=\frac{1}{c^2}-d^2=\frac{1}{c^2}-(\frac{1}{c}-\frac{1}{b}{)}^2=\frac{2}{bc}-\frac{1}{b^2}$

Next, $E=\frac{1}{c^2}-\frac{1}{4}(\frac{1}{c}-\frac{1}{a}{)}^2=\frac{1}{4}(\frac{3}{c^2}+\frac{2}{ca}-\frac{1}{a^2})$

and $E=(\frac{2}{b}-\frac{1}{a}{)}^2-(\frac{1}{b}-\frac{1}{a}{)}^2=\frac{3}{b^2}-\frac{2}{ab}$

Hence, options 'A', 'B' and 'C' are correct.