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

Given terms of A.P. 25, 22, 19, …………..

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

S_n = 116

First Term = a = 25, Common Difference = d = 22 – 19 = -3

Putting the value of a, d and S_n

⇒ 116 × 2 = n [50 – 3n + 3]

⇒ 232 = 53n – 3n²⇒ 3n² – 53n + 232 = 0

⇒ 3n² – 24n – 29n + 232 = 0

⇒ 3n (n – 8) – 29 (n – 8) = 0

⇒ (3n – 29) (n – 8) = 0

⇒ n = 8 or 29/3

n cannot be equal to 29/3. Therefore, n = 8

Last term = a₈ = a + (n – 1)d

a₈ = 25 + (8 – 1) × (-3) = 25 – 21

∴ a₈ = 4