Question

Find the value of k for which 8k + 4, 6k - 2 and 2k + 7 are three consecutive terms of an A.P.​

Answer 1

let the terms are in AP

hence it have common difference

d= (6k-2)-(8k+4) = 6k -2 -8k -4 = -2k-6

d= (2k+7)-(6k-2)= 2k+7 -6k+2 = -4k +9

hence

-2k -6 = -4k +9

4k-2k = 9+6

2k = 15

k= 15/2