$10 \cdot 46$ In a real geometric sequence, the sum of the first two terms is 7, the sum of the first six terms is 91, the sum of the first four terms is
(A) 28.
(B) 32.
(C) 35.
(D) 49.
(E) 84.
(32nd American High School Mathematics Examination, 1981)
[Solution] Let $a$ and $r$ be the first term and the common ratio of the sequence, respectively. Then we have
$$
\left\{\begin{array}{l}
a + a r = 7, \\
\frac{a\left(r^{6}-1\right)}{r-1} = 91.
\end{array}\right.
$$
(2) $\div$ (1) gives
$$
\frac{\frac{a\left(r^{6}-1\right)}{r-1}}{a(1+r)} = \frac{91}{7},
$$

which simplifies to $\frac{r^{6}-1}{r^{2}-1} = 13$, or $r^{4} + r^{2} - 12 = 0$, which can be factored as
$$
\left(r^{2}+4\right)\left(r^{2}-3\right) = 0.
$$

Since $r$ is a real number, then $r^{2} = 3$.
Therefore, $a + a r + a r^{2} + a r^{3}$
$$
\begin{array}{l}
= (a + a r) + r^{2}(a + a r) \\
= (a + a r)(1 + r^{2}) = 7 \cdot 4 = 28.
\end{array}
$$

Hence, the answer is $(A)$.