If $x, y, z$ are real and $9 x^{2} + 16 y^{2} + 25 z^{2} - 12 x y - 20 y z - 15 z x = 0,$ then $x, y, z$ are in

A.P.

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G.P.

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H.P.

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None of the above

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Solution

The correct option is C H.P.
$9 x^{2} + 16 y^{2} + 25 z^{2} - 12 x y - 20 y z - 15 z x = 0$
$\Rightarrow (3 x)^{2} + (4 y)^{2} + (5 z)^{2} - (3 x) (4 y) - (4 y) (5 z) - (5 z) (3 x) = 0$
It is of the form $a^{2} + b^{2} + c^{2} - a b - b c - a c = 0$
which can be written as
$\frac{1}{2} [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}] = 0$
$\Rightarrow a = b = c$

Given condition is
$(3 x - 4 y)^{2} + (4 y - 5 z)^{2} + (5 z - 3 x)^{2} = 0$
$\Rightarrow 3 x = 4 y = 5 z$
Let $3 x = 4 y = 5 z = k$
$\Rightarrow x = \frac{k}{3}, y = \frac{k}{4}, z = \frac{k}{5}$
$∵ \frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in A.P.
so, $x, y, z$ are in H.P.

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