Suppose $A, B, C$ are defined as $A = a^{2} b + a b^{2} - a^{2} c - a c^{2}, B = b^{2} c + b c^{2} - a^{2} b - a b^{2}$ and $C = a^{2} c + a c^{2} - b^{2} c - b c^{2}$ , where $a > b > c > 0$ and the equation $A x^{2} + B x + C = 0$ has equal roots, then $a, b, c$ are in

A . P .

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G . P .

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H . P .

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A . G . P .

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Solution

The correct option is B $H . P .$
$A = a^{2} b + a b^{2} - a^{2} c - a c^{2} = a (b - c) (a + b + c)$
$B = b^{2} c + b c^{2} - a^{2} b - a b^{2} = b (c - a) (a + b + c)$ , and
$C = a^{2} c + a c^{2} - b^{2} c - b c^{2} = c (a - b) (a + b + c)$
.
If $A x^{2} + B x + C = 0$ has equal roots, then
$B^{2} - 4 A C = 0$
$b^{2} (c - a)^{2} = 4 a c (b - c) (a - b)$
$b^{2} (a^{2} - 2 a c + c^{2}) = 4 a c (a b + b c - a c - b^{2})$
$a^{2} b^{2} - 2 a b^{2} c + b^{2} c^{2} = 4 a^{2} b c + 4 a b c^{2} - 4 a^{2} c^{2} - 4 a b^{2} c$
Simplify to get,
$(a b - 2 a c + b c)^{2} = 0$
$a b + b c = 2 a c$
Dividing by $a b c$
$\frac{1}{a} + \frac{1}{c} = \frac{2}{b}$
$∴$ $a, b, c$ are in H.P.

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