smc

Let $S$ be the first arithmetic sequence and $T$ be the second arithmetic sequence. If $n=1$, then $S_1:T_1=8:31$. Since $S_1$ and $T_1$ are just the first term, the first term of $S$ is $8a$ and the first term of $T$ is $31a$ for some $a$. If $n=2$, then $S_2:T_2=15:35=3:7$, so the sum of the first two terms of $S$ is $3b$ and the sum of the first two terms of $T$ is $7b$ for some $b$. Thus, the second term of $S$ is $3b-8a$ and the second term of $T$ is $7b-31a$, so the common difference of $S$ is $3b-16a$ and the common difference of $T$ is $7b-62a$. Thus, using the first terms and common differences, the sum of the first three terms of $S$ equals $\frac{1}{2}\cdot 3(16a+2(-16a+3b))$, and the sum of the first three terms of $T$ equals $\frac{1}{2}\cdot 3(62a+2(-62a+7b))$. That means $$\frac{16a+2(-16a+3b)}{62a+2(-62a+7b)}=\frac{22}{39}$$ $$\frac{6b-16a}{14b-62a}=\frac{22}{39}$$ $$\frac{3b-8a}{7b-31a}=\frac{22}{39}$$ $$117b-312a=154b-682a$$ $$-37b=-370a$$ $$b=10a$$ With the substitution, the common difference of $S$ is $14a$, and the common difference of $T$ is $8a$. That means the $11^{\text{th}}$ term of $S$ is $8a+10(14a)=148a$, and the $11^{\text{th}}$ term of $T$ is $31a+10(8a)=111a$. Thus, the ratio of the eleventh term of the first series to the eleventh term of the second series is $148a:111a=\boxed{4:3}$.