Question 640306
let the 1st no =x
2nd no =x+2
& 3rd no =x+4
the sum of their squares is 56 more than three times the square of the smallest number.
so x^2+(x+2)^2+(x+4)^2=56+3*x^2
or x^2+x^2+4x+4+x^2+8x+16=56+3x^2
or3x^2+12x+20=56+3x^2
or 12x=56-20=36
or x=36/12=3
so the nos are 3,5,7    ans