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Q: What integers are four consecutive integers whose sum is 114.?

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The numbers are 27, 28, 29 and 30.

118

The integers are 27, 28, 29 and 30.

27+28+29+30=114

27, 28, 29 and 30.

27, 28, 29 and 30

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

27 28 29 30

117

27 + 28 + 29 + 30 = 114

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 37, 38 and 39.

The numbers are 56 and 58.

114/4 = 28.5 So the 4 numbers are 27, 28, 29 and 30

27, 28, 29 and 30 is one set that meets the requirement. 37, 38 and 39 is another. There are many such sets.

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

The larger is 114, and the other is 112.

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

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 13 consecutive numbers from 114 to 126 are composite.The 13 consecutive numbers from 114 to 126 are composite.The 13 consecutive numbers from 114 to 126 are composite.The 13 consecutive numbers from 114 to 126 are composite.

The first ten consecutive composite positive integers are: 114 115 116 117 118 119 120 121 122 123

114.As we know that the numbers are 2 consecutive even integers, we know that one number will be 2 larger than the other. We can use this to solve the problem with algebra:Where x is the smaller number,x + (x + 2) = 2262x + 2 = 2262x = 224x = 112Therefore the larger integer is 112 + 2 = 114

37+38+39=114

114 115 116 117 118 119 120 121 122 123 124 125 126

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