1. **State the problem:** We are given the 15th term ($T_{15}$) and the 31st term ($T_{31}$) of an arithmetic series and need to find the first term ($a$) and common difference ($d$). Then, find the 100th term ($T_{100}$). 2. **Recall the formula for the $n$th term of an arithmetic series:** $$T_n = a + (n-1)d$$ 3. **Write equations for the given terms:** $$T_{15} = a + 14d = 43$$ $$T_{31} = a + 30d = 183$$ 4. **Subtract the first equation from the second to eliminate $a$:** $$a + 30d = 183$$ $$-(a + 14d = 43)$$ \begin{aligned} &\cancel{a} + 30d - \cancel{a} - 14d = 183 - 43 \\ &16d = 140 \end{aligned}$$ 5. **Solve for $d$:** $$d = \frac{140}{16} = \frac{35}{4} = 8.75$$ 6. **Substitute $d$ back into one of the original equations to find $a$:** $$a + 14 \times 8.75 = 43$$ $$a + 122.5 = 43$$ $$a = 43 - 122.5 = -79.5$$ 7. **Find the 100th term using the formula:** $$T_{100} = a + 99d = -79.5 + 99 \times 8.75$$ $$T_{100} = -79.5 + 866.25 = 786.75$$ **Final answers:** - First term $a = -79.5$ - Common difference $d = 8.75$ - 100th term $T_{100} = 786.75$