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