Given the sum of the certain number of terms of an A.P. = 116

We know that, S_n = \(\frac{n}{2}\)[2a + (n − 1)d]

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms So for the given A.P.(25, 22, 19,…)

Here we have, the first term (a) = 25

The sum of n terms S_n = 116

Common difference of the A.P. (d) = a₂ – a₁ = 22 – 25 = -3

Now, substituting values in S_n

⟹ 116 = \(\frac{n}{2}\)[2(25) + (n − 1)(−3)]

⟹ (\(\frac{n}{2}\))[50 + (−3n + 3)]  = 116

⟹ (\(\frac{n}{2}\))[53 − 3n] = 116

⟹ 53n – 3n² = 116 x 2

Thus, we get the following quadratic equation,

3n² – 53n + 232 = 0

By factorization method of solving, we have

⟹ 3n² – 24n – 29n + 232 = 0

⟹ 3n( n – 8 ) – 29 ( n – 8 ) = 0

⟹ (3n – 29)( n – 8 ) = 0

So, 3n – 29 = 0

⟹ n = 29/3

Also, n – 8 = 0

⟹ n = 8

Since, n cannot be a fraction, so the number of terms is taken as 8.

So, the term is:

a₈ = a₁ + 7d = 25 + 7(-3) = 25 – 21 = 4

Hence, the last term of the given A.P. such that the sum of the terms is 116 is 4.