2023 AMC 8 第 25 题

Let $d$ be the common difference

Use the widest possible bounds for $a_1$ and $a_{15}$ to force $d$

Let $d$ be the common difference. Using the largest possible value $10$ for $a_1$ and the smallest possible value $241$ for $a_{15},$ we have $$d \geq \dfrac{241 - 10}{14} = 16.5.$$ Since all the numbers are integers, $d$ must be at least $17.$

Using the smallest possible value $1$ for $a_1$ and the largest possible value $250$ for $a_{15},$ we have $$d \leq \dfrac{250 - 1}{14} \approx 17.8.$$ Since all the numbers are integers, $d$ is at most $17.$ Therefore, $d=17.$

Note that $17 \cdot 14 = 238.$ Since $a_{15}$ is at least $241,$ $a_1$ must be at least $3.$

On the other hand, if $a_1$ were greater than $3,$ then $a_2=a_1+17$ would be greater than $20,$ which is not allowed.

Now we know that $a_1 = 3$ and $d = 17.$ This tells us that $$a_{14} = 3 + 13 \cdot 17 = 224.$$

Therefore, sum of the digits is $$2 + 2 + 4 = 8.$$

Thus, A is the correct answer.