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.
$25 (9 x^{2} + y^{2}) + 9 z^{2} - 15 (5 x y + y z + 3 z x) = 0$
as we know $a^{2} + b^{2} + c^{2} - a b - b c - c a$
$= \frac{1}{2} ((a - b)^{2} + (b - c)^{2} + (c - a)^{2})$
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 $(15 x)^{2} + (5 y)^{2} + (3 z)^{2} - (15 x) (5 y) - (5 y) (3 z) - (3 z) (15 x) = 0$
$\Rightarrow (15 x - 5 y)^{2} + (5 y - 3 z)^{2} + (3 z - 15 x)^{2} = 0$
$\Rightarrow 15 x = 5 y = 3 z$
$\Rightarrow \frac{x}{1} = \frac{y}{3} = \frac{z}{5} = t$
$x = k, y = 3 k, z = 5 k$
$∴ x, y, z$ are in AP.

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