Given $S_n = S_{3n}$, can we express $a$ in terms of $n$?

Given $S_n = S_{3n}$, can we express $a$ in terms of $n$?

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 $n$.

Solution

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

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