1. **Problem statement:** We are given an arithmetic series where the 15th term is 43 and the 31st term is 183. We need to find: a) The first term ($a_1$) and the common difference ($d$). b) The 100th term of the series. 2. **Formula for the $n$th term of an arithmetic series:** $$a_n = a_1 + (n-1)d$$ where $a_n$ is the $n$th term, $a_1$ is the first term, and $d$ is the common difference. 3. **Set up equations using the given terms:** For the 15th term: $$a_{15} = a_1 + 14d = 43$$ For the 31st term: $$a_{31} = a_1 + 30d = 183$$ 4. **Solve the system of equations:** Subtract the first equation from the second: $$a_1 + 30d - (a_1 + 14d) = 183 - 43$$ $$a_1 + 30d - a_1 - 14d = 140$$ $$16d = 140$$ $$d = \frac{140}{16} = \frac{35}{4} = 8.75$$ 5. **Find the first term $a_1$:** Substitute $d = 8.75$ into the first equation: $$a_1 + 14 \times 8.75 = 43$$ $$a_1 + 122.5 = 43$$ $$a_1 = 43 - 122.5 = -79.5$$ 6. **Find the 100th term:** Use the formula: $$a_{100} = a_1 + 99d = -79.5 + 99 \times 8.75$$ $$a_{100} = -79.5 + 866.25 = 786.75$$ **Final answers:** - First term $a_1 = -79.5$ - Common difference $d = 8.75$ - 100th term $a_{100} = 786.75$