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

The reporter forgot to multiply the decimal by 100, making it 25%