If 1- b - a -1- b - c -1- a -1- c- then a - b - c- are inA- A-PB- G-PC- none of theseD- H-P

The correct option is D H.P 1/b a+1/b c=1/a+1/c 2b a c/(b a)(b c)=a+c/ac 2abc a^2c ac^2=(b^2 ab bc+ac)(a+c) 2abc a^2c ac^2=b^2a a^2b abc+a^2c+cb^2 abc bc^2+ ...

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Byju's Answer

Standard XI

Mathematics

AM,GM,HM Inequality

If 1/ b - a +...

Question

If $\frac{1}{b - a} + \frac{1}{b - c} = \frac{1}{a} + \frac{1}{c},$ then a,b,c, are in

A.P

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G.P

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H.P

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none of these

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Solution

The correct option is C
H.P

$\frac{1}{b - a} + \frac{1}{b - c} = \frac{1}{a} + \frac{1}{c}$

$\frac{2 b - a - c}{(b - a) (b - c)} = \frac{a + c}{a c}$

$2 a b c - a^{2} c - a c^{2} = (b^{2} - a b - b c + a c) (a + c)$

$2 a b c - a^{2} c - a c^{2} = b^{2} a - a^{2} b - a b c + a^{2} c + c b^{2} - a b c - b c^{2} + a c^{2}$

$∴ 4 a b c - 2 a^{2} c - 2 a c^{2} - b^{2} a + a^{2} b - c b^{2} + b c^{2} = 0$

$\Rightarrow (\frac{2}{b} - \frac{1}{a} - \frac{1}{c}) = 0$

$\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$ answer H.P

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