Question

Determine the value of k for which k² + 4k + 8, 2k² + 3k + 6 and 3k² + 4k + 4 are in A.P.

Answer 1

Since, (k² + 4k + 8), (2k² + 3k + 6) and (3k² + 4k + 4) are in A.P, we have

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

`=>` 2k² + 3k + 6 – k² – 4k – 8 = 3k² + 4k + 4 – 2k² – 3k – 6

`=>` k² – k – 2 = k² + k – 2

`=>` k² – k – k² – k = –2 + 2

`=>` –2k = 0

`=>` k = 0