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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