Determine k so that k^2 + 4k + 8,2k^2 + 3k + 6,3k^2 + 4k + 4 are three consecutive terms of an A.P.

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Solution

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

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