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

First term, a = 25 and

Common difference, d = 22 – 25 = -3

Also given, sum of certain number of terms of the A.P. is 116

The number of terms be n

So, we have

S_n = \( \dfrac{n}{2}\) [2a + (n-1)d] = 116

116 = \( \dfrac{n}{2}\) [2(25) + (n-1)(-3)]

116 x 2 = n [50 – 3n + 3]

232 = n [53 – 3n]

232 = 53n – 3n²

3n² – 53n + 232 = 0

3n² – 24n – 29n+ 232 = 0

3n(n – 8) – 29(n – 8) = 0

(3n – 29) (n – 8) = 0

Hence,

n =\( \dfrac{29}{3}\) or n = 8

As n can only be an integral value, n = 8

Thus, 8^th term is the last term of the A.P.

a₈ = 25 + (8 – 1)(-3)

= 25 – 21

= 4