For what value of k will the consecutive terms 2k+1 3k+3 and 5k-1form an ap
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==》3k+3-(2k+1)-(5k-1)= 0
==》3k+3-2k-1-5k+1=0
==》-4k+3=0
==》-4k=-3
==》4k=3
==》k=3/4
Thus it is the required value of k
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