VJesus12 wrote:If the average (arithmetic mean) of seven consecutive integers is k + 2, then the product of the greatest and least integer is

A. k^2 - 9
B. k^2 - 2k + 1
C. k^2 + 4k - 5
D. k^2 + 6k + 9
E. k^2 + 4k - 12

[spoiler]OA=C[/spoiler]

Source: Magoosh

We can let the smallest integer in the set = x and the largest = x + 6; thus:

(x + x + 6)/2 = k + 2

2x + 6 = 2k + 4

2x = 2k - 2

x = k - 1

Thus, the smallest integer is k - 1 and the greatest is k + 5, so the product of those two values is:

(k - 1)(k + 5) = k^2 + 4k - 5

Alternate solution:

Since k + 2 is in the average of the 7 consecutive integers, it's also the median, i.e., the 4th integer. Thus, the largest integer is 3 more than k + 2, i.e., k + 5, and the least integer is 3 less than k + 2, i.e., k - 1. The product of the greatest and least integer is:

(k + 5)(k - 1) = k^2 + 4k - 5

Answer: C