Q1 of 178 Page 551

Find the sum of each of the following APs:

(1/15), (1/12), (1/10),…….. to 11 terms.

Here, first term = 1/15

Common difference = (1/12) - (1/15) = 1/60


Sum of first n terms of an AP is


Sn = [2a + (n - 1)d]


∴ S11 = [2(1/15) + (11 - 1)(1/60)]


= (11/2) × [(2/15) + (1/6)]


= (11/2) × [(3/10)]


= 33/20


Thus, sum of 11 terms of this AP is 33/20.


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