Question 376038
find three consecutive integers such that the sum of the second and third is forty eight more than one-fifth the smallest.

Let x = the smallest integer
  x+1 = the 2nd integer
  x+2 = the 3rd integer

(x+1)+(x+2) = x/5 + 48
2x+3        = .2x +48
       1.8x = 45
          x = 25
       x+1  = 26
       x+2  = 27

The 3 consecutive integers are 25, 26, & 27.