Q. 114.2( 16 Votes )

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

Answer :

Let a_{1} = k^{2} + 4k + 8

a_{2} = 2k^{2} + 3k + 6

a_{3} = 3k^{2} + 4k + 4

Three terms will be in an AP if

a_{2} - a_{1} = a_{3} - a_{2}

2k^{2} + 3k + 6 - (k^{2} + 4k + 8) = 3k^{2} + 4k + 4 - (2k^{2} + 3k + 6)

k^{2} - k - 2 = k^{2} + k - 2

2k = 0

k = 0

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