The geometric progressions $a_{1}, a_{2}, a_{3}, \ldots$ and $b_{1}, b_{2}, b_{3}, \ldots$ have the same ratio, with $a_{1}=$ $27, b_{1}=99$ and $a_{15}=b_{11}$. Find the value of $a_{9}$.

#
Solution

Let $r$ be the common ratio of the two geometric progressions. We have

\[
\begin{aligned}
a_{15} & =a_{1} r^{14} \\
& =27 r^{14} \\
b_{11} & =b_{1} r^{10} \\
& =99 r^{10}
\end{aligned}
\]

Therefore, $27 r^{14}=99 r^{10}$ and thus $3 r^{4}=11$. Finally

\[
\begin{aligned}
a_{9} & =a_{1} r^{8} \\
& =27\left(r^{4}\right)^{2} \\
& =27\left(\frac{11}{3}\right)^{2} \\
& =\frac{27 \cdot 121}{9} \\
& =363
\end{aligned}
\]