In ΔABC, if a+b1−ab,b,b+c1−bc are in A.P, then a,1b,c are in

a+b1−ab,b,b+c1−bcare in A.P2b=a+b1−ab+b+c1−bc2b−a+b1−ab=b+c1−bc2b−2ab2−a−b1−ab=b+c1−bc⇒(2b−2ab2−a−b)(1−bc)=(b+c)(1−ab)⇒2b−2ab2−a−b−2b2c+2ab3c+abc+b2c=b+c−ab2−abc⇒−a+b−2ab2−b2c+2ab3c+abc=b+c−ab2−abc⇒−a−c−2ab2+ab2−b2c+2ab3c+abc+abc=0⇒−1(a+c)−b2(a+c)=−2abc(b2+1)⇒(b2+1)(a+c)=2abc(b2+1)⇒a+c2ac=b⇒1b=2aca+c⇒a,1b,c are in H.P

In ΔABC, if a+b1−ab,b,b+c1−bc are in A.P, then a,1b,c are in

a+b1−ab,b,b+c1−bcare in A.P2b=a+b1−ab+b+c1−bc2b−a+b1−ab=b+c1−bc2b−2ab2−a−b1−ab=b+c1−bc⇒(2b−2ab2−a−b)(1−bc)=(b+c)(1−ab)⇒2b−2ab2−a−b−2b2c+2ab3c+abc+b2c=b+c−ab2−abc⇒−a+b−2ab2−b2c+2ab3c+abc=b+c−ab2−abc⇒−a−c−2ab2+ab2−b2c+2ab3c+abc+abc=0⇒−1(a+c)−b2(a+c)=−2abc(b2+1)⇒(b2+1)(a+c)=2abc(b2+1)⇒a+c2ac=b⇒1b=2aca+c⇒a,1b,c are in H.P

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In $Δ A B C$ , if $\frac{a + b}{1 - a b}, b, \frac{b + c}{1 - b c}$ are in A.P, then $a, \frac{1}{b}, c$ are in

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Detailed Solution

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

$\begin{array}{l} 2 b = \frac{a + b}{1 - a b} + \frac{b + c}{1 - b c} \\ 2 b - \frac{a + b}{1 - a b} = \frac{b + c}{1 - b c} \\ \frac{2 b - 2 a b^{2} - a - b}{1 - a b} = \frac{b + c}{1 - b c} \end{array}$

$\begin{array}{l} \Rightarrow (2 b - 2 a b^{2} - a - b) (1 - b c) = (b + c) (1 - a b) \\ \Rightarrow 2 b - 2 a b^{2} - a - b - 2 b^{2} c + 2 a b^{3} c + a b c + b^{2} c \\ = b + c - a b^{2} - a b c \\ \Rightarrow - a + b - 2 a b^{2} - b^{2} c + 2 a b^{3} c + a b c = b + c - a b^{2} - a b c \end{array}$

$\begin{array}{l} \Rightarrow - a - c - 2 a b^{2} + a b^{2} - b^{2} c + 2 a b^{3} c + a b c + a b c = 0 \\ \Rightarrow - 1 (a + c) - b^{2} (a + c) = - 2 a b c (b^{2} + 1) \\ \Rightarrow (b^{2} + 1) (a + c) = 2 a b c (b^{2} + 1) \\ \Rightarrow \frac{a + c}{2 a c} = b \\ \Rightarrow \frac{1}{b} = \frac{2 a c}{a + c} \end{array}$