10.59 The sum of an infinite geometric series with common ratio $r$ (where $|r|<1$) is 15, and the sum of the squares of each term of this series is 45. The first term of this series is
(A) 12.
(B) 10.
(C) 5.
(D) 3.
(E) 2.
(21st American High School Mathematics Examination, 1970)
[Solution] Let the first term of the sequence be $a$, then we have
$$
\frac{a}{1-r}=15 \text {, and } \frac{a^{2}}{1-r^{2}}=45 \text {, }
$$

From $\frac{a^{2}}{1-r^{2}}=\frac{a}{1+r} \cdot \frac{a}{1-r}$, we get $\frac{a}{1+r}=3$. From $a=15(1-r)$ and $a=3(1+r)$, we can solve to get $a=5$. Therefore, the answer is $(C)$.