Gmat_mission wrote: ↑Thu May 21, 2020 1:45 am
The sum of all the integers \(k\) such that \(−26 < k < 24\) is
(A) 0
(B) −2
(C) −25
(D) −49
(E) −51
[spoiler]OA=D[/spoiler]
Source: Official Guide
Solution:
We must determine the sum of the consecutive integers from -25 to 23, inclusive. To determine the sum we can use the formula sum = average x quantity.
To determine quantity, the number of consecutive integers, we compute the following:
quantity = largest number – smallest number + 1
quantity = 23 – (-25) + 1 = 23 + 25 + 1 = 49
Next we must determine the average. Since we have a set of evenly-spaced integers we can determine the average using the formula:
average = (largest number + smallest number)/2.
average = (-25 + 23)/2 = -2/2 = -1
Finally we can determine the sum:
sum = quantity x average
sum = 49 x -1 = -49.
Alternate solution:
We must determine the sum of the consecutive integers from -25 to 23 inclusive. However, if we add -23 and 23, the sum will be 0, and so will be the sum of -22 and 22, -21 and 21, and so on. Therefore, the sum of each of these pairs of numbers (as long as they are opposites) is 0. The only numbers left that are not paired with their opposites are -25, -24 and 0. So the sum of all the integers from -25 to 23, inclusive, is the same as the sum of -25, -24 and 0, which is (-25) + (-24) + 0 = -49.
Answer: D
