Answer: D

Solution:

Write the $n$th term as $a+2n,$ so the term before the first term is $a.$ Then $$\begin{aligned}S_n&=\sum_{i=1}^n(a+2i)\\&=n(a+n+1).\end{aligned}$$

Hence $$\begin{aligned} \frac{S_{3n}}{S_n}&=\frac{3(a+3n+1)}{a+n+1}\\ &=9-\frac{6(a+1)}{a+n+1}. \end{aligned}$$

For this expression to be independent of $n,$ its numerator $6(a+1)$ in the final fraction must be $0.$ Thus $a=-1.$

Therefore, $$S_{20}=20(-1+20+1)=400.$$

Thus, the answer is D .