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

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(B) Given that $\frac{1}{b - c}, \frac{1}{c - a}, \frac{1}{a - b}$ are in $A.P.$Therefore,$2 	imes \frac{1}{c - a} = \frac{1}{b - c} + \frac{1}{a - b

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(B) Given that $\frac{1}{b - c}, \frac{1}{c - a}, \frac{1}{a - b}$ are in $A.P.$Therefore,$2 	imes \frac{1}{c - a} = \frac{1}{b - c} + \frac{1}{a - b}$.$\frac{2}{c - a} = \frac{(a - b) + (b - c)}{(b - c)(a - b)} = \frac{a - c}{(b - c)(a - b)}$.Since $a - c = -(c - a)$,we have $\frac{2}{c - a} = \frac{-(c - a)}{(b - c)(a - b)}$.$(c - a)^2 = -2(b - c)(a - b) = 2(b - c)(b - a)$.We want to check if $(b - c)^2, (c - a)^2, (a - b)^2$ are in $A.P.$This is true if $2(c - a)^2 = (b - c)^2 + (a - b)^2$.Substituting $(c - a)^2 = 2(b - c)(b - a)$,we get $4(b - c)(b - a) = (b - c)^2 + (a - b)^2$.$4(b - c)(b - a) - (b - c)^2 - (a - b)^2 = 0$.This simplifies to $-( (b - c) - (a - b) )^2 = 0$,which is not generally true.Wait,let us re-evaluate: The condition for $x, y, z$ to be in $A.P.$ is $2y = x + z$.Given $\frac{1}{b - c}, \frac{1}{c - a}, \frac{1}{a - b}$ are in $A.P.$Let $x = b - c, y = c - a, z = a - b$. Note $x + y + z = 0$.The given condition is $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in $A.P.$$\frac{2}{y} = \frac{1}{x} + \frac{1}{z} = \frac{x + z}{xz} = \frac{-y}{xz}$.So $y^2 = -2xz$.We want to check if $x^2, y^2, z^2$ are in $A.P.$This requires $2y^2 = x^2 + z^2$.Since $x + z = -y$,then $(x + z)^2 = (-y)^2$,so $x^2 + z^2 + 2xz = y^2$.$x^2 + z^2 = y^2 - 2xz$.Since $y^2 = -2xz$,then $x^2 + z^2 = y^2 - (-y^2) = 2y^2$.Thus,$x^2, y^2, z^2$ are in $A.P.$

If $\frac{1}{b - c}, \frac{1}{c - a}, \frac{1}{a - b}$ are consecutive terms of an $A.P.$,then $(b - c)^2, (c - a)^2, (a - b)^2$ will be in

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