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The integers are 37, 38 and 39.

Q: What is the sum of the three consecutive integers that equals 114?

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Since the average of the three integers will be 114/3 = 38, and the three numbers are consecutive, the numbers will be 36, 38 and 40. Another way to do this problem using algebra is to let the first integer be n, then the next two are n+2 and n+4. Their sum is 3n+6 and it equals 114 So 3n+6=114 and 3n=108 so n=36 then next two numbers must be 38 and 40 since they are consecutive even integers.

114/3 = 38, so the three integers are 37, 38 and 39.

27 + 28 + 29 + 30 = 114

118

Let the integers be x, (x+1) and (x+2) Since, x+(x+1)+(x+2) = -114, by solving the equation for x, we get that, x = -37 Hence the integers are -37, -36 and -35

Related questions

27+28+29+30=114

Since the average of the three integers will be 114/3 = 38, and the three numbers are consecutive, the numbers will be 36, 38 and 40. Another way to do this problem using algebra is to let the first integer be n, then the next two are n+2 and n+4. Their sum is 3n+6 and it equals 114 So 3n+6=114 and 3n=108 so n=36 then next two numbers must be 38 and 40 since they are consecutive even integers.

114/3 = 38, so the three integers are 37, 38 and 39.

The consecutive odd integers for 114 are 37, 38 and 39.

27 + 28 + 29 + 30 = 114

27 + 28 + 29 + 30 = 114

118

Let the integers be x, (x+1) and (x+2) Since, x+(x+1)+(x+2) = -114, by solving the equation for x, we get that, x = -37 Hence the integers are -37, -36 and -35

The integers are 27, 28, 29 and 30.

Three consecutive integers whose sum is 117 are 38, 39, and 40. N + (N+1) + (N+2) = 117 3N + 3 = 117 3N = 114 N = 38

27, 28, 29 and 30.

The numbers are 56 and 58.