Q: The sum of four consecutive negative integers?

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The sum is four.

The integers are 17, 19, 21 and 23.

No.Let the four numbers be (n-1), n, (n+1), (n+2).Their sum is 4n + 2 = 2(2n + 1)If 2000 is the sum of four consecutive integers, then:2(2n+1) = 2000⇒ 2n + 1 = 1000⇒ 2n = 999but 999 is odd, not even and so n cannot be an integer; therefore 2000 is not the sum of four consecutive integers.

The numbers are -11, -10, -9 and -8.

17 , 19 , 21 and 23 are the odd integers whose sum is 80 and the least integer is 17

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The sum of the first four non-negative, consecutive, even integers is 20.

The four consecutive integers with the sum of -66 are -18, -17, -16 and -15.

The two consecutive negative odd integers having 74 as the sum of their squares are -5 and -7.

-8, -9, -10, -11

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.

There is no set of three consecutive odd or even integers whose sum is negative 31.

They are odd consecutive integers: 21, 23, 25 and 27.