Question

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

Answer 1

Step-by-step explanation:

We know that three terms p,q,r form consecutive terms of AP if and only if 2q = p+r

Thus, 2k + 7, 6k - 2 and 8k + 4 will form consecutive terms of an AP is 2(6k-2) = (2k+7) + (8k+4)

Now, 2(6k-2) = (2k+7) + (8k+4)

12k - 4 = 10k + 11

2k = 15

k = 15/2