Let the first of the four consecutive integers be represented by n, then the other three can be (n + 1), (n + 2) and (n + 3)
Then we have, n + (n + 1) + (n + 2) + (n + 3) = -142
which simplifies to, 4n + 6 = -142
so, 4n = -142 - 6 = -148........n = -148/4 = -37
The four integers are thus, -37, -36 (-37 + 1), -35 (-37 + 2) and -34 (-37 + 3)
and we have, -37 +(-36) + (-35) + (-34) = -142
14, 15, 16 and 17.
There are four consecutive odd integers: 81, 83, 85 and 87.
There are no two consecutive whole numbers that sum to 142.With two consecutive whole numbers, one is even and the other odd, and their sum will be odd, but 142 is even.Two consecutive even numbers that sum to 142 are 70 and 72.
The integers are -92, -91, -90 and -89.
The integers are 51, 53, 55 and 57.
The sum of the first four non-negative, consecutive, even integers is 20.
The smallest is -1
If n is the smallest of the four integers, their sum is 4n+6.
No. The sum of four consecutive integers is two odd numbers plus two even numbers which is an even number. 2001 is an odd number, therefore it cannot be the sum of four consecutive numbers.
There are none.
There is a set of four consecutive even integers whose sum is four. The set is: -2, 0, 2 and 4.
14, 15, 16 and 17.
They are odd consecutive integers: 21, 23, 25 and 27.
6,7,8,9
The sum is four.
There are four consecutive odd integers: 81, 83, 85 and 87.
-4, -2, 0, and 2 are the four consecutive even integers. When you add them up they equal -4.