10.63 Let $S_{n}$ be the sum of the first $n$ terms of a harmonic series, for example, $S_{3}$ is the sum of the first three terms. If the first three terms of a harmonic series are $3,4,6$, then
(A) $S_{4}=20$.
(B) $S_{4}=25$.
(C) $S_{5}=49$.
(D) $S_{6}=49$.
(E) $S_{2}=\frac{1}{2} S_{4}$.
(10th American High School Mathematics Examination, 1959)
[Solution] Since $3,4,6$ form a harmonic sequence, then $\frac{1}{3}, \frac{1}{4}, \frac{1}{6}$ form an arithmetic sequence, with a common difference of $-\frac{1}{12}$. The fifth term of this sequence is $0 \left(\because \frac{1}{3}+(5-1) \cdot \frac{(-1)}{12}=0\right)$. Since the reciprocal of 0 does not exist, this harmonic sequence has only four terms.
Notice that $\frac{1}{6}-\frac{1}{12}=\frac{1}{12}$, so the fourth term of the harmonic sequence is 12.
Thus, $S_{4}=3+4+6+12=25$.
Therefore, the answer is $(B)$.