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

The correct option is C H.Pa + b/1 ab + b + c/1 bc = 2b ⇒ (a+b) (1 bc) + (b+c) (1 ab) = 2b (1 ab) (1 bc) ⇒ (b^2 + 1)(2abc (a+c)) = 0 (on simplification) ∴1/b ...

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Byju's Answer

Standard XII

Mathematics

Harmonic Progression

If 1/1-a b, b...

Question

$I f \frac{1}{1 - a b}, b, \frac{1}{1 - b c} a r e i n A . P . t h e n a, \frac{1}{b}, c a r e i n$

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
$\frac{a + b}{1 - a b} + \frac{b + c}{1 - b c} = 2 b \Rightarrow (a + b) (1 - b c) + (b + c) (1 - a b) = 2 b (1 - a b) (1 - b c) \Rightarrow (b^{2} + 1) (2 a b c - (a + c)) = 0 (o n s i m p l i f i c a t i o n) ∴ \frac{1}{b} = \frac{2 a c}{a + c}$

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If11−ab,b,11−bc are in A.P. then a,1b,c are in

The correct option is C H.Pa+b1−ab+b+c1−bc=2b⇒(a+b)(1−bc)+(b+c)(1−ab)=2b(1−ab)(1−bc)⇒(b2+1)(2abc−(a+c))=0 (on simplification)∴1b=2aca+c

$I f \frac{1}{1 - a b}, b, \frac{1}{1 - b c} a r e i n A . P . t h e n a, \frac{1}{b}, c a r e i n$

The correct option is C H.P
$\frac{a + b}{1 - a b} + \frac{b + c}{1 - b c} = 2 b \Rightarrow (a + b) (1 - b c) + (b + c) (1 - a b) = 2 b (1 - a b) (1 - b c) \Rightarrow (b^{2} + 1) (2 a b c - (a + c)) = 0 (o n s i m p l i f i c a t i o n) ∴ \frac{1}{b} = \frac{2 a c}{a + c}$