1. **Problem Statement:** Determine the sum of each arithmetic series given in question 8. 2. **Formula for the sum of an arithmetic series:** $$S_n = \frac{n}{2} (t_1 + t_n)$$ where $S_n$ is the sum of the first $n$ terms, $t_1$ is the first term, and $t_n$ is the $n$th term. 3. **Important rules:** - To find the sum, you need the number of terms $n$, the first term $t_1$, and the last term $t_n$. - If $t_n$ is not given, use the general term formula $t_n = t_1 + (n-1)d$ to find it. --- **a) $5 + 11 + 17 + ... + t_{21}$** - First term $t_1 = 5$ - Common difference $d = 11 - 5 = 6$ - Number of terms $n = 21$ - Last term $t_{21} = t_1 + (21-1) \times 6 = 5 + 20 \times 6 = 5 + 120 = 125$ - Sum: $$S_{21} = \frac{21}{2} (5 + 125) = \frac{21}{2} \times 130 = 21 \times 65 = 1365$$ **b) $4 - 1 - 6 - ... - t_{18}$** - First term $t_1 = 4$ - Common difference $d = -1 - 4 = -5$ - Number of terms $n = 18$ - Last term $t_{18} = 4 + (18-1)(-5) = 4 - 85 = -81$ - Sum: $$S_{18} = \frac{18}{2} (4 + (-81)) = 9 \times (-77) = -693$$ **c) $12 + 14 + 16 + ... + 68$** - First term $t_1 = 12$ - Common difference $d = 14 - 12 = 2$ - Last term $t_n = 68$ - Find $n$: $$68 = 12 + (n-1) \times 2 \Rightarrow 68 - 12 = 2(n-1) \Rightarrow 56 = 2(n-1) \Rightarrow n-1 = 28 \Rightarrow n = 29$$ - Sum: $$S_{29} = \frac{29}{2} (12 + 68) = \frac{29}{2} \times 80 = 29 \times 40 = 1160$$ **d) $14 + 11 + 8 + ... - 55$** - First term $t_1 = 14$ - Common difference $d = 11 - 14 = -3$ - Last term $t_n = -55$ - Find $n$: $$-55 = 14 + (n-1)(-3) \Rightarrow -55 - 14 = -3(n-1) \Rightarrow -69 = -3(n-1) \Rightarrow n-1 = 23 \Rightarrow n = 24$$ - Sum: $$S_{24} = \frac{24}{2} (14 + (-55)) = 12 \times (-41) = -492$$ **e) $t_1 = 4$ and $t_{20} = 144$** - Number of terms $n = 20$ - Sum: $$S_{20} = \frac{20}{2} (4 + 144) = 10 \times 148 = 1480$$ **f) $t_1 = -30$ and $t_{30} = 60$** - Number of terms $n = 30$ - Sum: $$S_{30} = \frac{30}{2} (-30 + 60) = 15 \times 30 = 450$$ **Final answers:** - a) 1365 - b) -693 - c) 1160 - d) -492 - e) 1480 - f) 450