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.