$10 \cdot 39$ Insert two positive numbers between 3 and 9 so that the first three numbers form a geometric sequence and the last three numbers form an arithmetic sequence. The sum of these two positive numbers is
(A) $13 \frac{1}{2}$.
(B) $11 \frac{1}{4}$.
(C) $10 \frac{1}{2}$.
(D) 10 .
(E) $9 \frac{1}{2}$.
(23rd American High School Mathematics Examination, 1972)
[Solution] Let $3, x, y$ be the first three numbers, then $\frac{x}{3}=\frac{y}{x}, x^{2}=3 y$. The last three numbers are $x, y, 9$, so $y-x=9-y, x+9=2 y$. Eliminating $y$ gives $2 x^{2}-3 x-27=0$, which is $(x+3)(2 x-9)=0$, solving for $x=4 \frac{1}{2}$ (negative value discarded), hence $y=\frac{x^{2}}{3}=6 \frac{3}{4}$, $\therefore x+y=11 \frac{1}{4}$.
Therefore, the answer is $(B)$.