Question

For what value of 'K' the numbers 3K + 2, 4K + 3 and 6K - 1 are the consecutive terms of an AP?

Answer:

Value of K is 5.

As 3K+2, 4K+3 and 6K-1 are in AP, (4K+3)-(3K+2) = (6K-1)-(4K+3) = common difference of the AP
Therefore, we have the equation:
4K+3-3K-2 = 6K-1-4K-3 which implies that K+1 = 2K-4 which further implies that K = 4+1 = 5.


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