If the n sum of a certain number of terms starting from first term of an A.P. is is 116. Find the last term.
a = 25, d = -3
Sn = [2a + (n – 1) d]
116 = [50 – 3n +3]
116 = [53 – 3n]
232 = 53n – 3n2
3n2 – 53n + 232 = 0
n = 8 or n = , which isn’t possible as n must be a natural number.
Therefore, n = 8
[a + l] = 116
[25 + l] = 116
l = 4
Hence, the last term is 4