1. **State the problem:** We are given the sum of the first $n$ terms of an arithmetic progression (AP) as $S_n = 3n^2 + 5n$. We need to find the $n$th term, $a_n$. 2. **Recall the formula:** The sum of the first $n$ terms of an AP is $S_n = \frac{n}{2}(2a_1 + (n-1)d)$, but here we have $S_n$ directly. The $n$th term can be found by the difference of sums: $a_n = S_n - S_{n-1}$. 3. **Calculate $a_n$:** $$a_n = S_n - S_{n-1} = (3n^2 + 5n) - [3(n-1)^2 + 5(n-1)]$$ 4. **Expand and simplify:** $$3(n-1)^2 = 3(n^2 - 2n + 1) = 3n^2 - 6n + 3$$ $$5(n-1) = 5n - 5$$ So, $$S_{n-1} = 3n^2 - 6n + 3 + 5n - 5 = 3n^2 - n - 2$$ 5. **Subtract:** $$a_n = (3n^2 + 5n) - (3n^2 - n - 2) = 3n^2 + 5n - 3n^2 + n + 2 = 6n + 2$$ 6. **Final answer:** The $n$th term of the AP is $$a_n = 6n + 2$$