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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$H.P.$

Vedclass · Answer

(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

Similar Questions

If $a, b, c$ are in $A.P.$,$b, c, d$ are in $G.P.$,and $c, d, e$ are in $H.P.$,then $a, c, e$ are in

If $9$ arithmetic means $(A.M.s)$ and $9$ harmonic means $(H.M.s)$ are inserted between $2$ and $3$,and if the harmonic mean $H$ corresponds to the arithmetic mean $A$ (i.e.,the $j^{th}$ $A.M.$ and $j^{th}$ $H.M.$),then $A + \frac{6}{H} = $

If the arithmetic mean of two positive numbers is $A$,their geometric mean is $G$,and their harmonic mean is $H$,then $H$ is equal to:

If $\alpha, \beta, \gamma$ are the geometric means between $ca, ab$; $ab, bc$; and $bc, ca$ respectively,where $a, b, c$ are in $A.P.$,then $\alpha^2, \beta^2, \gamma^2$ are in

If $a, b, c$ are in $A.P.$,then $2^{ax + 1}, 2^{bx + 1}, 2^{cx + 1}$ for $x \ne 0$ are in

Vedclass Test Series

Exam Paper Generator

Online Exam Module