if 2k+1,3k+3 and 5k-1 are in ap then find k
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since they are in AP then their common difference is same
d==(3k+3) - (2k+1) = (5k-1) -(3k+3)
==k+2 =2k-4
k = 6
d==(3k+3) - (2k+1) = (5k-1) -(3k+3)
==k+2 =2k-4
k = 6
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