If $x, y, z$ are real numbers satisfying the equation $25 (9 x^{2} + y^{2}) + 9 z^{2} - 15 (5 x y + y z + 3 z x) = 0$ then, $x, y, z$ 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 $A . P .$
We know that
$a^{2} + b^{2} + c^{2} - a b - b a - c a = \frac{1}{2} {{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}}$
So here
$25 ({9 x}^{2} + y^{2}) + 9 z^{2} - 15 (5 x y) + y z + 3 x z = 0$
can be arranged as
${(15 x)}^{2} + {(5 y)}^{2} + {(3 z)}^{2} - (15 x .5 y) - (5 y .3 z) - (3 z .15 x) = 0$
$\Rightarrow {(15 x - 5 y)}^{2} + {(5 y - 3 z)}^{2} + {(3 z - 15 x)}^{2} = 0$
$\Rightarrow 15 x = 5 y = 3 z$
$\Rightarrow \frac{x}{1} = \frac{y}{3} = \frac{z}{5} = k$ (say)
$\Rightarrow∴ x = k y = 3 k z = 5 k$
$\Rightarrow x, y, z$ are in $A P$
Option A.

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