1.

P2.For what value ofk will theFind the 9th term from the end (towards the fst terml of the A.P. 5,9,13.. . 185 (1531ltvide of k WIllK+9,2k -1 and 2k + 7 are the consecutive terms of an AP (k-18)consecutive terms 2k + 1, 3k + 3 and 5k-1 form an A.Prk-6)3.

Answer»

a2k + 1 = a3k + 3 = b5k - 1 = c

To form an AP,a + c = 2b2k + 1 + 5k - 1 = 2( 3k + 3)7k = 6k + 6k = 6

So, k = 6



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