We have $a_k$ = 5k + 1 By substituting the value of k = 1, 2, 3 and 4, we get $a_1$ = 5 + 1 = 6, $a_2$ = 10 + 1 = 11, $a_3$ = 15 + 1 = 16, $a_4$ = 20 + 1 = 21 Therefore the A.P. is 6, 11, 16, 21, ..... . Thus, a = 6, d = 11– 6 = 5 Hence, sum of n terms , $S_n$ = $\frac{n}{2}$ [2 (6) + (n - 1) 5] [Since $S_n$ = n/2 [2a + (n - 1) d] = $\frac{n}{2}$ [12 + 5n - 5] = $\frac{n}{2}$ [7 + 5n] = $\frac{5n^2+7n}{2}$