Question

please solve it fast and also give me steps

Answer 1

let the no be x,(x+2),(x+4)

so $x^{2}$+$(x+2)^{2}$+$(x+4)^{2}$=251$x^{2} +x^{2} +4x+8+x^{2} +8x+16=251[tex] x^{2} +x^{2} +4x+8+x^{2} +8x+16=251 3x^{2} +12x+20=251 3x^{2} +12x-231 solve the quadratic equation we get x=7 or x=-11 consider the positive value so x=7,x+2=9,x+4=11and their sum of squares =251$

Answer 2

let the nos be 2n+1,2n+3,2n+5
given that 3{(2n+1)^2+(2n+3)^2+(2n+5)^2}=251
12n^2+36n+35=251
12n^2+36n=216
n^2+3n=18
n^2+6n-3n-18=0
(n+6)(n-3)=0
so n=3,-6
I hope that it is clear 4 u