Let the sum of n terms of the given A.P. be 116.

Here, a = 25 and d = 22 – 25 = – 3

However,
n cannot be equal to 29/3 Therefore, n = 8
∴a8 = Last term = a + (n – 1)d = 25 + (8 – 1) (– 3)
= 25 + (7) (– 3) = 25 – 21
= 4
Thus, the last term of the A.P. is 4.