$34 \cdot 3$ Define $r!=r \cdot(r-1) \cdots \cdot 1$ and $C_{j}^{k}=\binom{j}{k}=\frac{j!}{k!(j-k)!}$, where $r, j, k$ are positive integers, and $k3$ when $C_{n}^{1}, C_{n}^{2}, C_{n}^{3}$ form an arithmetic sequence, then $n$ equals
(A) 5 .
(B) 7 .
(C) 9 .
(D) 11 .
(E) 12 .
(33rd American High School Mathematics Examination, 1982)
［Solution］From the problem, we have
$$
C_{n}^{1}=n, \quad C_{n}^{2}=\frac{n(n-1)}{2}, \quad C_{n}^{3}=\frac{n(n-1)(n-2)}{6} \text {. }
$$

And $\frac{n(n-1)}{2}-n=\frac{n(n-1)(n-2)}{6}-\frac{n(n-1)}{2}$.
After rearranging, we get
$$
n^{3}-9 n^{2}+14 n=0,
$$

which simplifies to
$$
n(n-2)(n-7)=0 \text {. }
$$

Since $n>3$, we have $n=7$.
Therefore, the answer is $(B)$.