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.

If x, y, z are real numbers satisfying the equation `25(9x^(2)+y^(2))+

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» 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