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

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(C) The given quadratic equation is $a(b - c)x^2 + b(c - a)x + c(a - b) = 0$.Since the roots are equal,the discriminant $D = 0$.Here,$A = a(b - c)$,$B

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$H.P.$

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(C) The given quadratic equation is $a(b - c)x^2 + b(c - a)x + c(a - b) = 0$.Since the roots are equal,the discriminant $D = 0$.Here,$A = a(b - c)$,$B = b(c - a)$,and $C = c(a - b)$.$B^2 - 4AC = 0 \Rightarrow [b(c - a)]^2 - 4[a(b - c)][c(a - b)] = 0$.$b^2(c - a)^2 - 4ac(b - c)(a - b) = 0$.$b^2(c^2 - 2ac + a^2) - 4ac(ab - b^2 - ac + bc) = 0$.$b^2c^2 - 2ab^2c + a^2b^2 - 4a^2bc + 4ab^2c + 4a^2c^2 - 4abc^2 = 0$.$b^2c^2 + 2ab^2c + a^2b^2 - 4a^2bc - 4abc^2 + 4a^2c^2 = 0$.$(bc + ab - 2ac)^2 = 0$.$bc + ab = 2ac$.Dividing by $abc$,we get $\frac{1}{a} + \frac{1}{c} = \frac{2}{b}$.This implies that $a, b, c$ are in $H.P.$

List-$I$	List-$II$
$(i)$ $\alpha = \beta$	$(A)$ $(ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$
$(ii)$ $\alpha = 2\beta$	$(B)$ $2b^2 = 9ac$
$(iii)$ $\alpha = 3\beta$	$(C)$ $b^2 = 6ac$
$(iv)$ $\alpha = \beta^2$	$(D)$ $3b^2 = 16ac$
	$(E)$ $b^2 = 4ac$
	$(F)$ $(ac^2)^{1/3} + (a^2c)^{1/3} = b$

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

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