Answer :

let three consecutive numbers be a-1 , a ,a+1

sum of above (a-1)+(a)+(a+1) = 237
                        3a = 237
                          a = 79

therefore three numbers are = 78, 79,80
Trim
Take the three consecutive numbers as x, x+1 and x+2.  These three have to add up to 237 so we get the equation:

x + (x+1) + (x+2) = 237
3x + 3 = 237
3x = 234
x = 234 / 3 = 78

Therefore the numbers are x (78), x+1 (79) and x+2 (80)

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