Let $S_n$ and $T_n$ be the respective sums of the first $n$ terms of two arithmetic series. If $S_n:T_n=(7n+1):(4n+27)$ for all $n$, the ratio of the eleventh term of the first series to the eleventh term of the second series is:
$\text{(A) } 4:3\quad \text{(B) } 3:2\quad \text{(C) } 7:4\quad \text{(D) } 78:71\quad \text{(E) undetermined}$
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 $\tfrac{1}{2} \cdot 3(16a + 2(-16a + 3b))$, and the sum of the first three terms of $T$ equals $\tfrac{1}{2} \cdot 3(62a + 2(-62a + 7b))$.  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 $148a:111a = \boxed{\textbf{(A) } 4:3}$.