Question

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

Answer 1

k²+4k+8 , 2k²+3k+6, 3k²+4k+4

are in AP

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

4k² + 6k + 12 = 4k² + 8k + 12

2k = 0

k = 0

hope it helps.