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# If $\large\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P., then the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ has

$\begin{array}{1 1} \text{(A) Real and distinct roots} \\ \text{(B) Real and equal roots} \\ \text{(C) No real roots } \\ \text{(D) cannot say } \end{array}$

Given $\large\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.
$\Rightarrow\: \large\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$.......$(i)$
Also since $a(b-c)x^2+b(c-a)x+c(a-b)=0$,
$x=1$ satisfy the equation.
$\therefore$ $1$ is one root and let the other root be $\alpha$
$\Rightarrow\:1.\alpha=\large\frac{c(a-b)}{a(b-c)}$
Dividing num. and den. by $abc$
$\alpha=\large\frac{\large\frac{1}{b}-\frac{1}{a}}{\large\frac{1}{c}-\frac{1}{b}}$
From $(i)\:\: \alpha=1$
$\therefore\:$ the equation has real equal roots = $1$