117, 118 and 119.
The numbers are 116, 118 and 120.
Let the four consecutive integers be ( x, x+1, x+2, x+3 ). The sum of these integers can be expressed as ( x + (x + 1) + (x + 2) + (x + 3) = 4x + 6 ). Setting this equal to -354 gives the equation ( 4x + 6 = -354 ). Solving for ( x ), we find ( x = -90 ), so the four consecutive integers are -90, -89, -88, and -87.
There is no set of three consecutive integers for 106.
The sum of the squares of the three consecutive integers 11, 13, 15 = 515
Three consecutive integers around 249 are 248, 249, and 250. Consecutive integers differ by one, so these numbers follow one another sequentially.
(-117)+(-118)+(-119)=(-354)
The numbers are 116, 118 and 120.
There is no set of three consecutive integers for 187.
Three consecutive integers have a sum of 12. What is the greatest of these integers?
Let the four consecutive integers be ( x, x+1, x+2, x+3 ). The sum of these integers can be expressed as ( x + (x + 1) + (x + 2) + (x + 3) = 4x + 6 ). Setting this equal to -354 gives the equation ( 4x + 6 = -354 ). Solving for ( x ), we find ( x = -90 ), so the four consecutive integers are -90, -89, -88, and -87.
There is no set of three consecutive integers for 106.
There is no set of three consecutive integers whose sum is 71.
The sum of three consecutive integers is -72
9240 is the product of the three consecutive integers 20, 21, and 22.
The sum of the squares of the three consecutive integers 11, 13, 15 = 515
There must be three consecutive integers to guarantee that the product will be divisible by 6. For the "Product of three consecutive integers..." see the Related Question below.
Three consecutive integers around 249 are 248, 249, and 250. Consecutive integers differ by one, so these numbers follow one another sequentially.