We have ak = 5k + 1
By substituting the value of k = 1, 2, 3 and 4, we get
a1 = 5 + 1 = 6, a2 = 10 + 1 = 11, a3 = 15 + 1 = 16, a4 = 20 + 1 = 21
Therefore the A.P. is 6, 11, 16, 21, ..... .
Thus, a = 6, d = 11– 6 = 5
Hence, sum of n terms , Sn =

n

2

[2 (6) + (n - 1) 5] [Since Sn = n/2 [2a + (n - 1) d]
=

n

2

[12 + 5n - 5] =

n

2

[7 + 5n] =

5n2+7n

2