# 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.

Sn = 116

First Term = a = 25, Common Difference = d = 22 – 19 = -3 Putting the value of a, d and Sn 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

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