Solution:

Let the sequence be $a_1, a_2, \cdots.$ Create $a$ by making it the term before $a_1$ in the sequence. This would make $a_n = a+2n.$

This would make $$S_n = \sum_{i=1}^n (a+2i) =$$ $$a\cdot n + 2\sum_{i=1}^ni = an + n^2+n.$$

This makes $$\dfrac{S_{3n}}{S_n} = \dfrac{3an + 9n^2+3n}{an + n^2+n} =$$ $$\dfrac{3a + 9n+3}{a + n+1}.$$ Thus, we must find $a$ such that this value is constant.

If our given value is constant, than the given value minus $9$ is constant, so $$\dfrac{3a + 9n+3}{a + n+1}-9 = \dfrac{-6a-6}{a+n+1}$$ is constant. As $n$ increases, the numerator is constant and the denominator is increasing. Therefore, if the number is constant, the numerator must be $0.$

Since $-6a-6 = 0,$ we have $a=-1.$

With our formula from before, we have $$S_{20} = -1(20)+20^2+20 = 400.$$

Thus, the answer is D .