If the roots of the equation a(b - c)x^2 + b( | vedclass

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(C) Given the quadratic equation $a(b - c)x^2 + b(c - a)x + c(a - b) = 0$.Since the sum of the coefficients is $a(b - c) + b(c - a) + c(a - b) = ab -

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$H.P.$

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(C) Given the quadratic equation $a(b - c)x^2 + b(c - a)x + c(a - b) = 0$.Since the sum of the coefficients is $a(b - c) + b(c - a) + c(a - b) = ab - ac + bc - ab + ac - bc = 0$,one root of the equation must be $1$.Since the roots are equal,both roots must be $1$.For a quadratic equation $Ax^2 + Bx + C = 0$,the product of roots is $\frac{C}{A}$.Thus,$1 	imes 1 = \frac{c(a - b)}{a(b - c)}$.$a(b - c) = c(a - b)$$ab - ac = ac - bc$$ab + bc = 2ac$Dividing both sides by $abc$,we get $\frac{1}{c} + \frac{1}{a} = \frac{2}{b}$.This implies that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$,which means $a, b, c$ are in $H.P.$

If the roots of the equation $a(b - c)x^2 + b(c - a)x + c(a - b) = 0$ are equal,then $a, b, c$ are in

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