The 104th term and 4th term of an A.P. are 125 and 0. Find the sum of first 35 terms.
Solution :
104th term = 125
4th term = 0
a + 103d = 125 -----(1)
a + 3d = 0 -----(2)
(1) - (2)
a + 103d - a - 3d = 125 - 0
100d = 125
d = 125/100 = 5/4
By applying the value of d in (2), we get
a + 3(5/4) = 0
a = -15/4
Sn = (n/2) [2a + (n - 1)d]
= (35/2) [2(-15/4) + (35 - 1)(5/4)]
= (35/2)[(-15/2) + (85/2)]
= (35/2)[70/2]
S35 = 612.5