Question

Find value of k for which k+2, 4k-6, 3k-2 form 3 consecutive terms of AP

Answer 1

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Answer 2

Given, these are the consecutive terms of a AP

Therefore, d1=d2

4k-6-(k+2) = 3k-2-(4k-6)

4k-6-k-2 = 3k-2-4k+6

4k-k-6-2 = 3k-4k-2+6

3k-8 = -k+4

3k+k = 4+8

4k = 12

k = 12/4

k = 3

Hence k=3 is the answer