In a sequence of 8 consecutive integers, how much greater is the sum

sum of first four integers=4x+6
sum of last four integers=4x+22
(4x+22)-(4x+6)=16
C

Or even simpler, realize that each of the 4 numbers is 4 more than its counterpart in the first half and there are 4 of them so the answer is 4*4

We can let the first 4 numbers be 1, 2, 3, and 4, and the last 4 be 5, 6, 7, and 8, thus the difference is:

26 - 10 = 16

Generally, we see that if the first 4 numbers are a, b, c, and d, the next 4 numbers will be (a + 4), (b + 4), (c + 4), and (d + 4). The sum of the first four numbers is (a + b + c+ d), and the sum of the next four numbers is (a + b + c + d + 16). Thus, the sum of the second group of four numbers is 16 greater than the sum of the first four numbers.

Answer: C

Hi Scott,

Thank you for your explanation - I also used this approach to solve it. Will this approach work on the more advanced versions of this problem as well?

Thanks!

Hi susheelh,

It is hard to give a definitive answer to that question; there will be instances where this approach will work, there will be instances where this approach will work with slight modifications, and there will be instances where this approach will not work at all.

For instance, the same method would work if you increased the number of consecutive integers (like 10 consecutive integers instead of 8). You could make it work by modifying the solution if you were looking at the difference between first n terms and last m terms (just represent all the integers as x, x + 1, x + 2 etc. instead of a, b, c, d).

It would also work for consecutive even/odd numbers, just make sure to determine the common difference correctly (in this question, if we had 8 consecutive even integers, we could use the a, b, c, d and a + 8, b + 8, c + 8, d + 8).

The first example that comes to mind in which your method wouldn't work is if, for example, we were looking at the difference between not the sum, but the product of the first so many numbers and the last so many numbers.

Posted from my mobile device

The \(5^{th}\) number will be 4 more than \(1^{st}\) number.
The \(6^{th}\) number will be 4 more than \(2^{nd}\) number.
The \(7^{th}\) number will be 4 more than \(3^{rd}\) number.
The \(8^{th}\) number will be 4 more than \(4^{th}\) number.


=> 4 + 4 + 4 + 4 or 4 * 4 = 16

Answer C