Answer: A

Video solution:

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Written solution:

Let $d$ be the common difference. If we let $a_1 = 10$ and $a_{15} = 241,$ we see that $$d \geq \dfrac{241 - 10}{14} = 16.5.$$ Since all the numbers are integers, $d$ must be at least $17.$

Also, if $a_1 = 1$ and $a_{15} = 250,$ we get that $$d \leq \dfrac{250 - 1}{14} \approx 17.8.$$ Once again since all the numbers are integers, $d$ is at most $17.$ This tells us that $d$ is $17.$

Note that $17 \cdot 14 = 238.$ This means that $a_1$ must be at least $3$ for $a_{15}$ to be within the desired range.

If $a_1$ is greater than $3,$ however, $a_2$ becomes 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.