1. **State the problem:** We are given the sum of a series from $n=3$ to $n=18$ of the terms defined by $6n - 10$. We need to find the first term ($a_1$), the last term ($a_{last}$), and the sum of the series. 2. **Identify the terms:** The general term is $a_n = 6n - 10$. 3. **Find the first term:** Since the series starts at $n=3$, the first term is $$a_3 = 6(3) - 10 = 18 - 10 = 8.$$ 4. **Find the last term:** The last term corresponds to $n=18$, so $$a_{18} = 6(18) - 10 = 108 - 10 = 98.$$ 5. **Number of terms:** The number of terms from $n=3$ to $n=18$ is $$18 - 3 + 1 = 16.$$ 6. **Sum of the series:** The sum of an arithmetic series is given by $$S = \frac{n}{2}(a_1 + a_n),$$ where $n$ is the number of terms, $a_1$ is the first term, and $a_n$ is the last term. 7. Substitute the values: $$S = \frac{16}{2}(8 + 98) = 8 \times 106 = 848.$$ **Final answers:** - First term $a_1 = 8$ - Last term $a_{last} = 98$ - Sum $S = 848$