14. Let $a_{1}, a_{2}, a_{3}, \cdots$ be a sequence of real numbers in a geometric progression. Let $S_{n}=$ $a_{1}+a_{2}+\cdots+a_{n}$ for all integers $n \geq 1$. Assume that $S_{10}=10$ and $S_{30}=70$. Find the value of $S_{40}$.
14. Answer: 150
Solution. Let $r$ be the common ratio of this geometric sequence. Thus
$$
S_{n}=a_{1}\left(1+r+r^{2}+\cdots+r^{n-1}\right) \text {. }
$$

Thus
$$
10=a_{1}\left(1+r+\cdots+r^{9}\right)
$$
and
$$
70=a_{1}\left(1+r+\cdots+r^{29}\right) .
$$
$\mathrm{As}$
$$
1+r+\cdots+r^{29}=\left(1+r+\cdots+r^{9}\right)\left(1+r^{10}+r^{20}\right),
$$
we have
$$
1+r^{10}+r^{20}=7 .
$$

So $r^{10}$ is either 2 or -3 . As $r^{10}>0, r^{10}=2$.
Hence
$$
\begin{array}{c}
S_{40}=a_{1}\left(1+r+\cdots+r^{39}\right)=a_{1}\left(1+r+\cdots+r^{9}\right)\left(1+r^{10}+r^{20}+r^{30}\right) \\
=10 \times\left(1+2+2^{2}+2^{3}\right)=150 .
\end{array}
$$