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Sagot :
let the first number be x, the second number be x+1, and the third number be x+2.
x + x+1 + x+2 = 114
3x + 3 = 114
3x = 111
x = 37
x+1 = 38
x+2 = 39
So, the three consecutive numbers are 37, 38, and 39.
x + x+1 + x+2 = 114
3x + 3 = 114
3x = 111
x = 37
x+1 = 38
x+2 = 39
So, the three consecutive numbers are 37, 38, and 39.
consecutive integers are represented by
first number=x
second=x+1
third=x+2
so x+x+1+x+2=114
solve
3x+3=114
subtract 3
3x=111
divide 3
37=x
the numbers are
37,38,39
first number=x
second=x+1
third=x+2
so x+x+1+x+2=114
solve
3x+3=114
subtract 3
3x=111
divide 3
37=x
the numbers are
37,38,39
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