SajjadAhmad wrote:

In a decreasing sequence of seven consecutive even integers, the sum of the first four integers is 68. What is the product of the last three integers in the sequence?

A. 1,000

B. 960

C. 925

D. 30

E. 25

The seven even integers in the sequence are as follows:

x, x - 2, x - 4, x - 6, x - 8, x - 10, x - 12

Since the sum of the first 4 integers is 68, we have:

x + x - 2 + x - 4 + x - 6 = 68

4x = 80

x = 20

Since our last 3 integers are x - 8, x - 10, and x - 12, they are 12, 10, and 8, and thus their product is 12 x 10 x 8 = 960.

Answer: B

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