If the sum of a certain number of terms of the A.P. 25, 22, 19,….. is 116, find the last term.

Here Solution = 116, a = 25, d = 22 – 25 = -3

Let n be the number of terms.

i.e., Last term = t_n

Now S_n =

or 116 =

or 232 = n[50 – 3n + 3]

or n[53 – 3n] = 232

or 3n² – 53n + 232 = 0

or 3n² – 29n – 24n + 232 = 0

or n(3n – 29) –8(3n – 29) =0

or (3n – 29) (n – 8) =0

or n = 8 as n ∑ N

Now Last term = T_n = a+(n – 1) d

or T₈ = 25 + (8 – 1)(-3)

        = 25 – 21 = 4

Therefore the last term is 4.