determine K so that 3K - 2, 4 k - 6 and K +2 are three consecutive terms of an AP
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5
If 3k-2,4k-6 and k+2 are in ap then therefore common difference between two consecutive terms will be equal. So that means 4k-6-(3k-2)=k+2-(4k-6) and when you solve the bracket stuff you get k-4=-3k+8 and then 4k=12 so k=3
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2
C d = a2 - a 1
K = 4
K = 4
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