Find the sum of first 28 terms of an A.P. whose nth term is 4n -3.

Find the sum of first 28 terms of an A.P. whose nth term is 4n -3.
Solution :
tn  = 4n - 3
n = sum of 28 terms  = 28
n = 1
tn  = 4n - 3
t1  = 4(1) - 3  = 4 - 3
tn  = 1  ==> a = 1
n = 2
tn  = 4n - 3
t2  = 4(2) - 3  = 8 - 3
t2  = 5 
d = 5 - 1 = 4
Sn  = (n/2) [2a + (n - 1)d]
  =  (28/2) [2(1) + (28 - 1)4]
  =  14[2 + 27(4)]
  =  14 [110]
  =  1540
Hence the sum of 28 terms of the given sequence is 1540.