Arithmetic Sequence Sum 741A7E

1. **Stating the problem:** We need to find the sum of the first 17 terms of the sequence starting with $-\frac{5}{9}$, $-\frac{11}{18}$, $-\frac{2}{3}$, and so on. 2. **Identify the pattern:** Let's check if this is an arithmetic sequence by finding the common difference $d$. Calculate $d = a_2 - a_1 = -\frac{11}{18} - \left(-\frac{5}{9}\right) = -\frac{11}{18} + \frac{5}{9}$. Convert $\frac{5}{9}$ to eighteenth: $\frac{5}{9} = \frac{10}{18}$. So, $d = -\frac{11}{18} + \frac{10}{18} = -\frac{1}{18}$. Check $a_3 - a_2 = -\frac{2}{3} - \left(-\frac{11}{18}\right) = -\frac{2}{3} + \frac{11}{18}$. Convert $-\frac{2}{3}$ to eighteenth: $-\frac{2}{3} = -\frac{12}{18}$. So, $a_3 - a_2 = -\frac{12}{18} + \frac{11}{18} = -\frac{1}{18}$. Since the difference is constant, the sequence is arithmetic with common difference $d = -\frac{1}{18}$. 3. **Formula for the sum of the first $n$ terms of an arithmetic sequence:** $$ S_n = \frac{n}{2} \left(2a_1 + (n-1)d\right) $$ Where: - $n = 17$ - $a_1 = -\frac{5}{9}$ - $d = -\frac{1}{18}$ 4. **Calculate the sum:** Calculate $2a_1 = 2 \times \left(-\frac{5}{9}\right) = -\frac{10}{9}$. Calculate $(n-1)d = 16 \times \left(-\frac{1}{18}\right) = -\frac{16}{18} = -\frac{8}{9}$. Sum inside parentheses: $$ 2a_1 + (n-1)d = -\frac{10}{9} - \frac{8}{9} = -\frac{18}{9} = -2 $$ Now calculate $S_{17}$: $$ S_{17} = \frac{17}{2} \times (-2) = 17 \times (-1) = -17 $$ 5. **Final answer:** The sum of the first 17 terms is **$-17$**.