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