JAMB 2016 · UTME · Q16

The sum of the first $n$ terms of the arithmetic progression $5, 11, 17, 23, 29, 35, \ldots$ is

Worked solution (try it first)

This is an A.P. with $a = 5$ and $d = 6$ .
Use $S_n = \frac n2\big(2a + (n - 1)d\big)$ : the bracket is $10 + 6(n - 1) = 6n + 4$ .
So $S_n = \frac n2(6n + 4) = n(3n + 2)$ , option B.