If the sum of $n$ terms of an A.P. is $2n+3n^2$, find the $r^{th}$ term. If the sum of $n$ terms of an Arithmetic Progression is $2n+3n^2$, find the $r^{th}$ term.
Note: This question is from the book  Higher Algebra by H.S. Hall & S.R. Knight and its answer is $6r-1$. (As listed at the end of the book).
What I've already done:
$$
S_n = 2n+3n^2\\
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}\\2n+3n^2=\frac{n}{2}\left \{ 2a+(n-1)d \right \}
$$

We know that the sum of the first $n$ terms is $S_n = 2n + 3n^2$, so the sum of the first $(n - 1)$ terms is $S_{n-1} = 2(n - 1) + 3(n - 1)^2$. But if the $n$th term is $t_n$, then recall that $S_{n-1} + t_n = S_n$. Hence, it follows that:
\begin{align*}
t_n
&= S_n - S_{n-1} \\
&= (2n + 3n^2) - (2(n - 1) + 3(n - 1)^2) \\
&= (2n + 3n^2) - (2n - 2 + 3(n^2 - 2n + 1)) \\
&= (2n + 3n^2) - (2n - 2 + 3n^2 - 6n + 3) \\
&= (2n + 3n^2) - (3n^2 - 4n + 1) \\
&= 6n - 1 \\
\end{align*}
as desired.

Remark: Notice that the formula for $S_n$ gives us a lot of information. Indeed, we didn't need to know that the sequence was an arithmetic progression, and we didn't need to memorize any fancy formulas.
