Geometric Sequence - Find the tenth term 
Find the $10$th term of a geometric sequence whose fifth element is $81$ and whose ninth element is $16.$

I have obtained a common ratio of $\tfrac23$ and my $a_1$ is $410.0625$, thus my $a_{10}$ is $10.66667$. Is this correct? Thank you!
 A: There are two such (real-valued) sequences. One is $a_n=\frac{3^9}{2^5} \left(\frac23\right)^n$ and the other one is $a_n=-\frac{3^9}{2^5} \left(-\frac23\right)^n$. You can obtain them by setting up the system $$\begin{cases}\alpha x^5=81\\ \alpha x^9=16\end{cases}$$ which yields $x^4=\frac{16}{81}$. Two real solutions for $x$, and thus two real values of $\alpha$. Of course, this provides two possible values of $a_{10}$.
A: Edited to include the fact that the ratio could be negative.
It is (partially) correct. One can also solve that problem without calculating $a_1$. In fact, you know $a_9 = r^4a_5$ so
$$r = \sqrt[4]{\frac{16}{81}}=\pm\frac23$$
And hence $a_{10} = ra_9 = \pm\frac{2}{3}16 \approx \pm10.66667$
A: The short way:
$$ar^{10}=\frac{(ar^9)^{5/4}}{(ar^5)^{1/4}}=\frac{16^{5/4}}{81^{1/4}}=\frac{32}3.$$

How was this found ? By working on the exponents of $a$ and $r$, we solve the following system:
$$\begin{cases}\lambda+5\mu=1,\\\lambda+9\mu=10\end{cases}.$$
Then $\lambda=-\dfrac14,\mu=\dfrac54$.
