If x2-9 y2-25 z2-xyz15-x-5-y-3-z then

The correct options are A x,y and z are in H.P. B 1x,1y,1z are in A.P.x^2+9y^2+25z^2=xyz (15x+5y+3z) ⇒ x^2+(3y)^2+(5z)^2=(3y)(5z)+x(5z)+x(3y) ⇒ x^2+(3y)^2+(5z)^ ...

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Byju's Answer

Standard XI

Mathematics

Inequality

If x2+9 y2+25...

Question

If $x^{2} + 9 y^{2} + 25 z^{2} = x y z (\frac{15}{x} + \frac{5}{y} + \frac{3}{z})$ then,

x, y

and

z

are in

H . P .

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\frac{1}{x}, \frac{1}{y}, \frac{1}{z}

are in

A . P .

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x, y

and

z

are in

G . P .

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\frac{1}{x}, \frac{1}{y}, \frac{1}{z}

are in

G . P .

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Solution

The correct options are
A $x, y$ and $z$ are in $H . P .$
B $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in $A . P .$
$x^{2} + 9 y^{2} + 25 z^{2} = x y z (\frac{15}{x} + \frac{5}{y} + \frac{3}{z})$
$\Rightarrow x^{2} + (3 y)^{2} + (5 z)^{2} = (3 y) (5 z) + x (5 z) + x (3 y)$
$\Rightarrow x^{2} + (3 y)^{2} + (5 z)^{2} - (3 y) (5 z) - x (5 z) - x (3 y) = 0 \Rightarrow \frac{(x - 3 y)^{2} + (3 y - 5 z)^{2} + (x - 5 z)^{2}}{2} = 0$
$\Rightarrow x = 3 y = 5 z = k \Rightarrow x = k, y = \frac{k}{3}, z = \frac{k}{5}$
$\Rightarrow \frac{1}{x} = \frac{1}{k}; \frac{1}{y} = \frac{3}{k}; \frac{1}{z} = \frac{5}{k}$
$x, y, z$ are in $H . P .$

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Similar questions

Q. If

x^{2} + 9 y^{2} + 25 z^{2} = x y z (\frac{15}{x} + \frac{5}{y} + \frac{3}{z})

then x, y, z are in

Q. If

x^{2} + 9 y^{2} + 25 z^{2} = x y z (\frac{15}{x} + \frac{5}{y} + \frac{3}{z})

where x, y, z

\in R

, then x, y, z satisfy.

Q. Assertion :If

x^{2} + 9 y^{2} + 25 z^{2} = x y z (\frac{15}{x} + \frac{5}{y} + \frac{3}{z})

, then x, y, z are in H.P. Reason: If

a_{1}^{2} + a_{2}^{2} + \dots + a_{n}^{2} = 0

, then

a_{1} = a_{2} = a_{3} = \dots = a_{n} = 0

Q. Simplify:

(x - 3 y - 5 z) (x^{2} + 9 y^{2} + 25 z^{2} + 3 x y - 15 y z + 5 z x)

Q. If

x^{2} + 9 y^{2} - 4 x + 3 = 0, x, y \in R,

then

x

and

y

respectively lie in the intervals

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If x2+9y2+25z2=xyz(15x+5y+3z) then,

The correct options are A x,y and z are in H.P. B 1x,1y,1z are in A.P.x2+9y2+25z2=xyz(15x+5y+3z) ⇒x2+(3y)2+(5z)2=(3y)(5z)+x(5z)+x(3y) ⇒x2+(3y)2+(5z)2−(3y)(5z)−x(5z)−x(3y)=0⇒(x−3y)2+(3y−5z)2+(x−5z)22=0 ⇒x=3y=5z=k⇒x=k,y=k3,z=k5 ⇒1x=1k;1y=3k;1z=5k x,y,z are in H.P.

If $x^{2} + 9 y^{2} + 25 z^{2} = x y z (\frac{15}{x} + \frac{5}{y} + \frac{3}{z})$ then,