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Question

If a, b, c are real numbers satisfying the equation 25(9a2+b2)+9c215(5ab+bc+3ca)=0 then prove that a, b, c are in A.P.

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Solution

Let x=5.3a,y=5b,z=3c
then x2+y2+z2xyyzzx=0
12[(xy)2+(yz)2+(zx)2]=0
xy=0,yz=0,zx=0
x=y=z=λ, say
15a=5b=3c=15k, say
a=k,b=3k,c=5k.

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