If $(k + 1), 3 k, (4 k + 2)$ are three consecutive terms of an AP, then k = ?
(a) 0
(b) 1
(c) 2
(d) 3

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Solution

(d) 3
If (k+1),3k and(4k+2) are three consecutive terms of an AP, then we have:
3k - (k + 1) = (4k + 2) - 3k
⇒ 2k - 1 = k +2
⇒ k = 3

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