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Mathematics Question on Sequence and series

If a+b1−ab,b,b−c1−bc\frac{a+b}{1-ab},b,\frac{b-c}{1-bc} are in A.P.., then a,1b,ca,\frac{1}{b},c are in

A

A.P.

B

G.P.

C

H.P.

D

none of these

Answer

H.P.

Explanation

Solution

By the given condition b−a+b1−ab=b+c1−ab−b b-\frac{a+b}{1-ab} = \frac{b+c}{1-ab} - b ⇒b−ab2−a−b1−ab=b+c−b+b2c1−bc\Rightarrow \frac{b-ab^{2}-a-b}{1-ab} =\frac{ b+c-b+b^{2}c}{1-bc} ⇒−a(1+b2)1−ab=c(1+b2)1−bc \Rightarrow \frac{-a\left(1+b^{2}\right)}{1-ab} = \frac{c\left(1+b^{2}\right)}{1-bc} ⇒−a1−ab=c1−bc\Rightarrow \frac{-a}{1-ab} = \frac{c}{1-bc} ⇒−a+abc=c−abc\Rightarrow -a+abc = c-abc ⇒2abc=c+a\Rightarrow 2abc= c+a ⇒1b=2aca+c\Rightarrow \frac{1}{b} = \frac{2ac}{a+c} ⇒a,1b,c \Rightarrow a, \frac{1}{b}, c are in H.PH.P.