1. Stating the problem: The first term ($a_1$) of an arithmetic series is 51, and the eighth term ($a_8$) is 100. We need to find the twentieth term ($a_{20}$). 2. Recall the general formula for the $n$-th term of an arithmetic series: $$a_n = a_1 + (n-1)d$$ where $d$ is the common difference. 3. Use the information for the eighth term to find $d$: $$a_8 = a_1 + 7d = 100$$ Plugging in $a_1 = 51$, we get $$51 + 7d = 100$$ 4. Solve for $d$: $$7d = 100 - 51$$ $$7d = 49$$ $$d = \frac{49}{7} = 7$$ 5. Find the twentieth term using the formula: $$a_{20} = a_1 + 19d = 51 + 19 \times 7 = 51 + 133 = 184$$ Final answer: $a_{20} = 184$ which corresponds to option a.