1.

If x, y, z are real numbers satisfying the equation 25(9x^(2)+y^(2))+9z^(2)-15(5xy+yz+3zx)=0 , then prove that x, y and z are in A.P.

Answer»

Solution :We hare,
`25(9x^(2)+y^(2))+9z^(2)-15(5xy+yz+3zx)=0`
`rArr (15X)^(2)+(5y)^(2)+(3z)^(2)-(15x)(5y)-(5y)(3z)-(3z)(15x)=0`
`rArr (1)/(2)[2(15x)^(2)+2(5y)^(2)+2(3z)^(2)-2(15x)(5y)-2(5y)(3z)-2(3z)(15x)]=0`
`rArr (1)/(2)[(15x-5y)^(2)+(5y-3z)^(2)+(3z-15x)^(2)]=0`
`rArr (15x-5y)^(2)+(5y-3z)^(2)+(3z-15x)^(2)=0`
It is possible only when
`15x-5y=0 and 5y-3z=0 and 3z-15x=0`
`rArr 15x=5y=3z`
`rArr (x)/(1)=(y)/(3)=(Z)/(5)=K` (say)
`:. x=k, y=3k, z=5k`
Now, `y-x=3k-k=2k`
and`z-y=5k-3k=2k`
Since, `y-x=z-y`
`:. x,y,z ` are in A.P. `""` Hence Proved


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