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Find the value of k for which 2k + 7, 6k - 2 and 8k + 4 form 3

consecutive terms of an AP.

Asked by Topperlearning User 4th June 2014, 1:23 PM
Answered by Expert
Answer:

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 =

Answered by Expert 4th June 2014, 3:23 PM
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