$10 \cdot 45$ Given a geometric sequence with the first term not equal to 0 and the common ratio not equal to 0, and an arithmetic sequence with the first term equal to 0. A third sequence is formed by adding the corresponding terms of the first two sequences, which is 1, 1, $2, \cdots$ The sum of the first ten terms of this third sequence is
(A) 978.
(B) 557.
(C) 467.
(D) 1068.
(E) Cannot be determined from the given information.
(6th American High School Mathematics Examination, 1955)
[Solution] Let the geometric sequence be $a, a q, a q^{2}, \cdots$ and the arithmetic sequence be
$$
0, \quad d, 2 d,
$$
$\qquad$
then the third sequence is $a, a q+d, a q^{2}+2 d$, $\qquad$
Given $a=1, a q+d=1, a q^{2}+2 d=2$.
From $q+d=1, q^{2}+2 d=2$ and $q \neq 0$, we solve to get
$$
q=2, \quad d=-1 .
$$

Therefore,
$$
\begin{array}{l}
S^{\prime}{ }_{10}=\frac{a\left(q^{10}-1\right)}{q-1}=1023, \\
S^{\prime \prime}{ }_{10}=\frac{10}{2}[0+(10-1) d]=-45,
\end{array}
$$
$$
S_{10}=1023-45=978 .
$$

Hence, the answer is $(A)$.