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Question

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

A
A.P.
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B
G.P.
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C
A.G.P.
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D
H.P.
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Solution

The correct option is D H.P.
Given, x2+9y2+25z2=15yz+5zx+3xy
(x)2+(3y)2+(5z)2(x)(3y)(3y)(5z)(x)(5z)=0
12[2(x)2+2(3y)2+2(5z)22(x)(3y)2(3y)(5z)2(x)(5z)]=0
12[(x3y)2+(3y5z)2+(x5z)2]=0
x3y=0,3y5z=0,x5z=0
1x=13yand53y=1zand15z=1x
1x+1z=13y+53y=2y
1x,1y,1z
are in A.P. and x,y,z are in H.P.

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