$20 \cdot 118$ A circle with area $A_{1}$ is contained within a larger circle with area $A_{1}+A_{2}$. If the radius of the larger circle is 3, and $A_{1}, A_{2}, A_{1}+A_{2}$ form an arithmetic sequence, then the radius of the smaller circle is
(A) $\frac{\sqrt{3}}{2}$.
(B) 1 .
(C) $\frac{2}{\sqrt{3}}$.
(D) $\frac{3}{2}$.
(E) $\sqrt{3}$.
(30th American High School Mathematics Examination, 1979)
［Solution］Since $A_{1}, A_{2}, A_{1}+A_{2}$ form an arithmetic sequence, then

i.e., $\square$
$$
\begin{aligned}
A_{2}-A_{1} & =\left(A_{1}+A_{2}\right)-A_{2}, \\
2 A_{1} & =A_{2} .
\end{aligned}
$$

If $r$ is the radius of the smaller circle, then,
$$
S_{\text {large circle }}=3^{2} \pi=9 \pi=A_{1}+A_{2}=3 A_{1}=3 \pi r^{2} \text {, then } r=\sqrt{3} \text {. }
$$
Therefore, the answer is $(E)$.