Question 125486
If the first integer is x, then the second must be x + 1, the third x + 2, and the fourth x + 3.  Their sum must then be {{{x+(x+1)+(x+2)+(x+3)}}}.  This sum is one more than the third.  The third consecutive integer is x + 2, so one more than that is x + 3.  Therefore we can write:


{{{x+(x+1)+(x+2)+(x+3)=x+3}}}


Removing the parentheses and collecting terms gets us to:
{{{4x+6=x+3}}}


Add -x to both sides and add -6 to both sides:
{{{3x=-3}}}


Divide by 3
{{{x=-1}}}


Therefore the integers are -1, 0, 1, and 2.


Check:  -1 + 0 + 1 + 2 = 2, so the sum of the integers is 2.
The third is 1, and one more than that is 2.  Answer checks.