# Determine k, so that k2 + 4k + 8, 2k2 + 3k + 6 and 3k2 + 4k + 4 are three consecutive terms of an AP.

Let a1 = k2 + 4k + 8

a2 = 2k2 + 3k + 6

a3 = 3k2 + 4k + 4

Three terms will be in an AP if

a2 - a1 = a3 - a2

2k2 + 3k + 6 - (k2 + 4k + 8) = 3k2 + 4k + 4 - (2k2 + 3k + 6)

k2 - k - 2 = k2 + k - 2

2k = 0

k = 0

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