Given the sum of the certain number of terms of an A.P. = 116
We know that, Sn = \(\frac{n}{2}\)[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 Sn = 116
Common difference of the A.P. (d) = a2 – a1 = 22 – 25 = -3
Now, substituting values in Sn
⟹ 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.