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

G . P .

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A . P .

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H . P .

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A . G . P .

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Solution

The correct option is C $H . P .$
$4 a^{2} + 9 b^{2} + 16 c^{2} = 2 (3 a b + 6 b c + 4 c a)$
$\Rightarrow (2 a)^{2} + (3 b)^{2} + (4 c)^{2} - (2 a) (3 b) - (3 b) (4 c) - (4 c) (2 a) = 0 \Rightarrow 2 (2 a)^{2} + 2 (3 b)^{2} + 2 (4 c)^{2} - 2 (2 a) (3 b) - 2 (3 b) (4 c) - 2 (4 c) (2 a) = 0$
$\Rightarrow (2 a - 3 b)^{2} + (3 b - 4 c)^{2} + (4 c - 2 a)^{2} = 0 \Rightarrow 2 a - 3 b = 0, 3 b - 4 c = 0, 4 c - 2 a = 0 ∴ 2 a = 3 b = 4 c = k (say)$
$\Rightarrow a = \frac{k}{2}, b = \frac{k}{3}, c = \frac{k}{4}$
So, $a, b, c$ are in $H . P .$

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