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Question

For what value of k will the consecutive terms 2K+1, 3K+3 and 5k-1 form an A. P? 


Solution

2k + 1 = a
3k + 3 = b
5k - 1 = c

To form an AP,
a + c = 2b
2k + 1 + 5k - 1 = 2( 3k + 3)
7k = 6k + 6
k = 6

So, k = 6

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