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 \(n\).

For an arithmetic progression, the sum of the first \(n\) terms, \(S_n = \dfrac{n}{2}(2a+(n-1)d)\).

So here, \(S_n = \dfrac{n}{2}(2a-n+1)\), and \(S_{3n} = \dfrac{3n}{2}(2a-3n+1)\).

Equating these gives \[\begin{align*} \dfrac{n}{2}(2a-n+1) &= \dfrac{3n}{2}(2a-3n+1)\\ \implies\quad 2a-n+1 &= 6a-9n+3\\ \implies\quad 4a &= 8n-2\\ \implies\quad a &= 2n - \dfrac{1}{2}. \end{align*}\]