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

The correct option is C H.P1/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+a ...

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Byju's Answer

Standard IX

Mathematics

Harmonic Progression

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

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

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

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

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

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

$\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$

$∴$ It is in H.P.

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The correct option is C H.P1b−a+1b−c=1a+1c 2b−a−c(b−a)(b−c)=a+cac 2abc−a2c−ac2=(b2−ab−bc+ac)(a+c) 2abc−a2c−ac2=b2a−a2b−abc+a2c+cb2−abc−bc2+ac2 ∴ 4abc−2a2c−2ac2−b2a+a2b−cb2+bc2=0 ⇒(2b−1a−1c)=0 2b=1a+1c ∴ It is in H.P.

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