Bunuel wrote:

In an increasing sequence of 8 consecutive even integers, the sum of the first 4 integers is 268. What is the sum of all the integers in the sequence?

A. 552

B. 568

C. 574

D. 586

E. 590

We can label the integers as follows:

x = first integer

x + 2 = second integer

x + 4 = third integer

x + 6 = fourth integer

x + 8 = fifth integer

x + 10 = sixth integer

x + 12 = seventh integer

x + 14 = eighth integer

Since the sum of the first 4 integers is 268:

x + x + 2 + x + 4 + x + 6 = 268

4x + 12 = 268

4x = 256

x = 64

So, we see that the first integer is 64 and the last integer is 14 + 64 = 78. Since we have an evenly spaced set of integers, the average of the set is (64 + 78)/2 = 142/2 = 71.

Since we have 8 integers, the sum is 8 x 71 = 568.

Alternative solution:

Like the previous solution, we can let x, x + 2, x + 4, x + 6, x + 8, x + 10, x + 12, and x + 14 be the 8 even consecutive integers. Notice that x, x + 2, x + 4, and x + 6 are the first 4 integers and x + 8, x + 10, x + 12, and x + 14 are the last 4 integers. Also notice that each of the last 4 integers is 8 more than each of the first 4 integers, correspondingly; thus, the sum of the last 4 integers is 8 x 4 = 32 more than the sum of the first 4 integers. Since we are given that the first 4 integers sum to 268, the last 4 integers sum to 268 + 32 = 300, and hence the 8 integers sum to 268 + 300 = 568.

Answer: B

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