The 104th term and 4th term of an A.P. are 125 and 0. Find the sum of first 35 terms.

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