Simple arithmetic progressions question

An arithmetic progression has first term a and common difference -1. The sum of the first n terms is equal to the sum of the first 3n terms. Express a in terms of n.

The answer is given and is 2n - 1/2

How do you do this?

Recall that s_n = (n/2)(a + t_n) = (n/2)(2a + d(n - 1)).

Thus, we know that:
s_n = s_[3n]
(n/2)(2a + d(n - 1)) = (3n/2)(2a + d(3n - 1))
(2a + d(n - 1)) = 3(2a + d(3n - 1)) <--[Divide both sides by n/2.]
2a + d(n - 1) = 6a + 3d(3n - 1) <--[Expand.]
2a - (n - 1) = 6a - 3(3n - 1) <--[Since d = -1.]
2a - n + 1 = 6a - 9n + 3 <--[Expand.]
(-n + 9n) + (1 - 3) = (6a - 2a) <--[Group like terms.]
8n - 2 = 4a
a = (8n - 2)/4 = 2n - 1/2, as desired.