If frac{a + b}{1 - ab}, b, frac{b + c}{1 - bc | vedclass

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(C) Given that $\frac{a + b}{1 - ab}, b, \frac{b + c}{1 - bc}$ are in $A.P.$This implies $b - \frac{a + b}{1 - ab} = \frac{b + c}{1 - bc} - b$$\Righta

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$H.P.$

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(C) Given that $\frac{a + b}{1 - ab}, b, \frac{b + c}{1 - bc}$ are in $A.P.$This implies $b - \frac{a + b}{1 - ab} = \frac{b + c}{1 - bc} - b$$\Rightarrow \frac{b(1 - ab) - (a + b)}{1 - ab} = \frac{(b + c) - b(1 - bc)}{1 - bc}$$\Rightarrow \frac{b - ab^2 - a - b}{1 - ab} = \frac{b + c - b + bc^2}{1 - bc}$$\Rightarrow \frac{-a(1 + b^2)}{1 - ab} = \frac{c(1 + b^2)}{1 - bc}$Dividing both sides by $(1 + b^2)$,we get $\frac{-a}{1 - ab} = \frac{c}{1 - bc}$$\Rightarrow -a(1 - bc) = c(1 - ab)$$\Rightarrow -a + abc = c - abc$$\Rightarrow 2abc = a + c$Dividing both sides by $abc$,we get $2b = \frac{1}{c} + \frac{1}{a}$This is the condition for $\frac{1}{a}, b, \frac{1}{c}$ to be in $A.P.$,which means $a, \frac{1}{b}, c$ are in $H.P.$

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

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