i dont now
There is no solution.Consecutive negative integers will always add up to a negative number. 110 is positive.
Consecutive negative integers that sum to 440 would be integers that are sequentially negative and add up to that positive number. For example, the integers -1, -2, -3, and so forth are negative integers, but their sum cannot reach 440 since they are all negative. If you meant the consecutive negative integers that multiply to give -440, those could be -20 and -22, as they are consecutive and their product is 440.
No two consecutive integers can add up to 98.No three consecutive integers can add up to 98.But 23, 24, 25, and 26 can.
n(n+2)(n+4) = 24
There are no such integers. Proof: choose any positive whole number x two consecutive even integers: (2x) (2x+2) Take the sum (2x)+(2x+2) (2x)+(2x+2)=340 4x+2=340 4x=338 x=338/4=84.5 Since this is not a whole number, there is no whole number that satisfies the conditions. (There are two consecutive odd integers which add up to 340: 169 and 171)
-4, -2, 0, and 2 are the four consecutive even integers. When you add them up they equal -4.
Let the two consecutive odd integers be ( x ) and ( x + 2 ). The equation can be set up as ( x + (x + 2) = 116 ). Solving for ( x ), we get ( 2x + 2 = 116 ), which simplifies to ( 2x = 114 ) and ( x = 57 ). Therefore, the two consecutive odd integers are 57 and 59.
10 12 14 are the consecutive even integers which add up to 36
Let the three consecutive integers be ( x ), ( x+1 ), and ( x+2 ). The equation can be set up as ( x + (x + 1) + (x + 2) = 252 ). Simplifying this gives ( 3x + 3 = 252 ), which leads to ( 3x = 249 ) and ( x = 83 ). Therefore, the three consecutive integers are 83, 84, and 85.
That isn't possible. The three consecutive number are assumed to be integers; the sum of three consecutive integers is always a multiple of 3 (try it out).
The integers are -11, -9 and -7.
The sum of the first 60 positive integers (1 + 2 + 3 + .... + 59 + 60) is equal to 1830.