Answer :
[tex]a,a+1,a+2,a+3-\ four\ consecutive\ integers\\\\
a+a+1+a+2+a+3=682\\\\
4a+6=682\ \ \ | subtract\ 6\\\\
4a=676\ \ \ | divide\ by\ 4\\\\
a=169\\\\
Number\ are\ 169,170,171,172.[/tex]
Because they're consecutive, it means that the first one is x, the second one is x+1, the third one is x+2 (x+1+1), and the fourth one is x+3. Because we're given that the sum of all of them is 682, just add all of them together:
x+(x+1)+(x+2)+(x+3)=682
x+x+x+x+1+2+3=682
4x+6=682
4x+6-6=682-6
4x=676
x=676/4
x=169
Thus, the consecutive integers are: 169, 170, 171, 172
x+(x+1)+(x+2)+(x+3)=682
x+x+x+x+1+2+3=682
4x+6=682
4x+6-6=682-6
4x=676
x=676/4
x=169
Thus, the consecutive integers are: 169, 170, 171, 172