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
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The integers are 37, 38 and 39.
27 + 28 + 29 + 30 = 114
118
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
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.
The integers are 37, 38 and 39.
The consecutive odd integers for 114 are 37, 38 and 39.
27 + 28 + 29 + 30 = 114
Three consecutive integers have a sum of 12. What is the greatest of these integers?
There is no set of three consecutive integers whose sum is 71.
118
The sum of three consecutive integers is -72
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
The integers are 27, 28, 29 and 30.
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.
27+28+29+30=114
The numbers are 56 and 58.