If $4 a^{2} + 9 b^{2} + 16 c^{2} = 4 (3 a b + 6 b c + 4 c a)$ , where a, b, c are non zero numbers, then a, b, c are in

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none of these

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Solution

The correct option is C HP
$4 a^{2} + 9 b^{2} + 16 c^{2} = 4 (3 a b + 6 b c + 4 c a)$
$\Rightarrow (4 a^{2} - 12 a b + 9 b^{2}) + (9 b^{2} - 24 b c + 16 c^{2}) + (16 c^{2} - 16 c a + 4 a^{2}) = 0$
$\Rightarrow {(2 a - 3 b)}^{2} + {(3 b - 4 c)}^{2} + {(4 c - 2 a)}^{2} = 0$
Therefore, $2 a - 3 b = 0$ ; $3 b - 4 c = 0$ ; $4 c - 2 a = 0$
$\Rightarrow 2 a = 3 b = 4 c = k$
$\Rightarrow a = \frac{k}{2}$ ; $b = \frac{k}{3}$ ; $c = \frac{k}{4}$
now $\frac{2}{b} = \frac{6}{k}$ and $\frac{1}{c} + \frac{1}{a} = \frac{6}{k}$
$\frac{1}{c} + \frac{1}{a} = \frac{2}{b}$
Therefore, $a, b, c$ are in H.P

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