Question

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

Answer 1

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Answer 2

Answer:

since they are consecutive,

a₂-a₁ = a₃-a₂

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

2k²+3k+6-k²-4k-8k= 3k²+4k+4-2k²-3k-6

⇒k²-k-2=k²+1k-2

⇒-k=1k

0=1k+k

2k=0

k=0