$10 \cdot 62$ The sequence $a_{1}, a_{2}, a_{3}, \cdots$ satisfies $a_{1}=19, a_{9}=99$ and for all $n \geqslant 3$, $a_{n}$ is the arithmetic mean of its preceding $n-1$ terms. Then $a_{2}$ is
(A) 29 .
(B) 59 .
(C) 79 .
(D) 99 .
(E) 179 .
(50th American High School Mathematics Examination, 1999)
[Solution] Given $a_{1}=19$, let $a_{2}=x$, thus the terms of this sequence are (first find the 3rd term, then find $a_{4}$ based on the 2nd and 3rd terms, and so on):
$19, x, \frac{19+x}{2}, \frac{3}{2}(19+x), \cdots$

It is easy to observe that starting from $a_{4}$, they are respectively multiplied by
$$
\begin{array}{l}
\left(1+\frac{1}{2}\right) \frac{1}{3},\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \frac{1}{4}, \\
\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right) \frac{1}{5}, \cdots
\end{array}
$$

Thus, $a_{9}=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \cdots\left(1+\frac{1}{7}\right) \cdot \frac{1}{8} \cdot(19+x)=99$, solving for $x$ gives $x=179$, i.e., $a_{2}=179$. Therefore, the answer is $(E)$.