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 Jun, 2014, 01:23: PM
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
Answered by | 4th Jun, 2014, 03:23: PM
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