Question

8k+4,6k-2,2k-7 fine the determine of k

Answer 1

Let 8k+4, 6k-2 and 2k-7 be 3 consecutive terms of an AP.

= (6K-2)-(8K+4) = (2K-7) - (6K-2)

= 6K-2-8K-4= 2K-7-6K +2

= -2K-6 = -4K-5

= 2K = 1

= K = 1/2