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Question
Determine k so that (3k -2), (4k – 6) and (k +2) are three consecutive terms of an AP.
Solution
It is given that (3k -2) ,(4k -6) and (k +2) are three consecutive terms of an AP.
∴ (4k - 6) - (3k - 2) = (k+2) - (4k - 6)
⇒ 4k - 6 - 3k + 2 = k+2 - 4k +6
⇒ k - 4 = -3k + 8
⇒ k+ 3k = 8+4
⇒ 4k = 12
⇒ k = 3
Hence, the value of k is 3.
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