Find four consequtive integers such that the latgest is 2 more than half the sum of the first threr integers

Consecutive integers are 1 apart

they are x, x+1, x+2, x+3

the largest is x+3

the largest is 2 more than half the sumo f the first 3

x+3=2 + (1/2) (x+x+1+x+2)

solve

x+3=2 + (1/2) (3x+3)

minus 2 both sides

x+1 = (1/2) (3x+3)

time both sides by 2

2x+2=3x+3

minus 2x both sides

2=x+3

minus 3 both sides

-1=x

x=-1

x+1=0

x+2=1

x+3=2

the integers are - 1,0,1,2