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Question: If \(\frac { a + b } { 1 - a b }\) , b , \(\frac { b + c } { 1 - b c }\) are in A.P. then a<sup>–1<...

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

A

A.P.

B

G.P.

C

H.P.

D

None of these

Answer

A.P.

Explanation

Solution

2b = a+b1−ab\frac { a + b } { 1 - a b } + b+c1−bc\frac { b + c } { 1 - b c }

̃ 2ab3 c + 2abc = a + c + ab2 + b2c

̃ 2abc (b2 +1) = (a + c) (1 + b2)

̃ 2b = a+cac\frac { \mathrm { a } + \mathrm { c } } { \mathrm { ac } } ̃ 2b = 1a\frac { 1 } { \mathrm { a } } +

̃ a–1, b, c–1 are in A.P.