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

Question

Philoid · Accepted Answer

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

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