Question

What is the sum of a 58-term arithmetic sequence where the first term is 6 and the last term is 405?

Answer 1

\[6 + 57d = 405\]\[d = \frac{405 - 6}{57} = 7\]\[S_n = \frac{n}{2}\left(2a_1 + (n-1)d\right)\]\[S_{58} = \frac{58}{2}\left(2(6) + (58-1)7\right) = 11919\]