Question

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

Answer 1

Since, k² + 4k + 8, 2k² + 3k + 6 and 3k² + 4k + 4 are consecutive terms of an AP.

∴ 2k² + 3k + 6 – (k² + 4k + 8)

= 3k² + 4k + 4 – (2k² + 3k + 6) ...[Common difference]

⇒ 2k² + 3k + 6 – k² – 4k – 8 = 3k² + 4k + 4 – 2k² – 3k – 6

⇒ k² – k – 2 = k² + k – 2

⇒ – k = k

⇒ 2k = 0

⇒ k = 0