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