Question

determine k so that 4k+8 2k2+3k+6 3k2+4k+4 are three consecutive terms of an ap​

Answer 1

Answer:

K=0

Step-by-step explanation:

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