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If a, b, c are in H.P., then \(\bigg(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\bigg)\)\(\bigg(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\bigg)\) is equal to(a) \(\frac{4}{b^2}-\frac{3}{ac}\)(b) \(\frac{3}{b^2}-\frac{4}{ac}\)(c) \(\frac{4}{ac}-\frac{3}{b^2}\)(d) \(\frac{3}{b^2}+\frac{4}{ac}\)

Answer»

(c) \(\frac{4}{ac}-\frac{3}{b^2}\)

If a, b, c are in H.P, then b = \(\frac{2ac}{a+c}\)

∴ \(\bigg(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\bigg)\)\(\bigg(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\bigg)\)

\(\bigg(\frac{1}{a}+\frac{a+c}{2ac}-\frac{1}{c}\bigg)\)\(\bigg(\frac{a+c}{2ac}+\frac{1}{c}-\frac{1}{a}\bigg)\)

\(\bigg(\frac{2c+a+c-2a}{2ac}\bigg)\)\(\bigg(\frac{a+c+2a-2c}{2ac}\bigg)\)

\(\bigg(\frac{3c-a}{2ac}\bigg)\)\(\bigg(\frac{3a-c}{2ac}\bigg)\) = \(\frac{10ac-3a^2-3c^3}{4a^2c^2}\)

\(\frac{16ac-(3a^2+3c^2+6ac)}{4a^2c^2}\) = \(\frac{16ac-3(a+b)^2}{4a^2c^2}\)

\(\frac{4}{ac}-3\bigg(\frac{a+c}{2ac}\bigg)^2\) = \(\frac{4}{ac}-\frac{3}{b^2}\)



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