10-29 Let $S_{n}$ and $T_{n}$ be the sums of the first $n$ terms of two arithmetic sequences, respectively. If for all $n$, $\frac{S_{n}}{T_{n}}=\frac{7 n+1}{4 n+27}$, then the ratio of the 11th term of the first sequence to the 11th term of the second sequence is
(A) $\frac{4}{3}$.
(B) $\frac{3}{2}$.
(C) $\frac{7}{4}$.
(D) $\frac{78}{71}$.
(E) cannot be determined.
(20th American High School Mathematics Examination, 1969)
[Solution] Let $S_{n}, T_{n}$ have the first terms and common differences as $a_{1}, a_{2}; d_{1}, d_{2}$, and let $u_{n}, t_{n}$ be the general terms of the two sequences, then
$$
S_{n}=n\left(a_{1}+\frac{n-1}{2} d_{1}\right), \quad T_{n}=n\left(a_{2}+\frac{n-1}{2} d_{2}\right) .
$$

Therefore, $\quad \frac{S_{n}}{T_{n}}=\frac{2 a_{1}+(n-1) d_{1}}{2 a_{2}+(n-1) d_{2}}=\frac{7 n+1}{4 n+27}$.
Also, $\quad \frac{u_{11}}{t_{11}}=\frac{a_{1}+10 d_{1}}{a_{2}+10 d_{2}}=\frac{2 a_{1}+20 d_{1}}{2 a_{2}+20 d_{2}}$,
By comparing with the above, we can see that when $n=21$, it is exactly the latter formula.
So $\frac{u_{11}}{t_{11}}=\frac{7 \cdot 21+1}{4 \cdot 21+27}=\frac{148}{111}=\frac{4}{3}$.
Therefore, the answer is $(A)$.