We can solve this problem using sum of n terms of an A.P. We know that tn = Sn−Sn−1 By using this formula we can find the 12th term We have The sum of n term of a series is
n(n+1)(n+2)
3
Take the given sum as Sn T12 = S12−S11 [Since tn = Sn−Sn−1] here n = 12 =
12(12+1)(12+2)
3
-
11(11+1)(11+2)
3
= 4 × 13 × 14 - 11 × 4 × 13 = 728 - 572 = 156 ∴ The sum of n terms of a series is