The three numbers are 24, 25, 26. You can solve this as a system of 3 equations and 3 unknowns. Let A, B, & C be the 3 numbers, so:
[A + B - C = 23] ; then two equations to describe they are consecutive:
[B - A = 1] & [C - B = 1]. Solving gives A=24, B=25, & C=26
If the numbers are consecutive then the middle number must be one-third of 51. The numbers are thus 16, 17 and 18.
Their sum is 99.
Call the numbers 2n, 2n+2, and 2n+4 for some integer n. Using the form 2n ensures the numbers are even and adding 2 to the first one to get the second and then adding 2 to the third one to get the third ensures they are consecutive. Their sum is 54 so 6n+6=54 -> 6n=48 and n=8 The numbers are: 16, 18, and 20
110, 111 and 112. The middle one must be a third of the total.
the IQR is the third quartile minus the first quartile.
The three consecutive whole numbers you are looking for are 1, 2, and 3. The sum of the first two numbers, 1 + 2 = 3.
121
5
If the numbers are consecutive then the middle number must be one-third of 51. The numbers are thus 16, 17 and 18.
No. Every third consecutive natural number is divisible by 3.
Their sum is 99.
The numbers are 29, 30, & 31. 29 * 4 = 116 ....... 31 * 3 = 93; 93 + 23 = 116
If the first number is 915, then the second will be 915 + 1 = 916 and the third number will be 915 + 2 = 917. Add 915, 916 and 917 to find the sum. If the question is, what are the three consecutive numbers whose sum is 915, then divide 915 by 3 to find the middle number which is 305. Since the consecutive numbers differ by 1, the numbers are 304, 305, and 306.
Call the numbers 2n, 2n+2, and 2n+4 for some integer n. Using the form 2n ensures the numbers are even and adding 2 to the first one to get the second and then adding 2 to the third one to get the third ensures they are consecutive. Their sum is 54 so 6n+6=54 -> 6n=48 and n=8 The numbers are: 16, 18, and 20
the sum of three numbers is 14.the first number minus three times the third number is the second number.the second number is six more than the first number.find the three numbers
110, 111 and 112. The middle one must be a third of the total.
You can find your first of these three numbers by expressing the question like this: x + (x + 2) = 128 ∴ 2x + 2 = 128 ∴ x + 1 = 64 ∴ x = 63 So your numbers are 63, 64 and 65.