Answer
Geometric;
$7(781+156\sqrt 5)$
Work Step by Step
Let's note by $\{a_n\}$ the given sequence. The sum of its first $9$ terms is:
$$S_9=\sum_{k=0}^8 7(5)^{k/2}=\sum_{n=1}^9a_n.$$
The terms of the sequence are:
$$\begin{align*}
a_1&=7(5)^{0/2}=7\\
a_2&=7(5)^{1/2}=7\sqrt 5\\
a_3&=7(5)^{2/2}=35\\
&\dots\\
a_9&=7(5)^{8/2}.
\end{align*}$$
We notice that the common ratio of consecutive terms is $\sqrt 5$, therefore constant, so the sequence is geometric with the elements:
$$\begin{cases}
a_1=7\\
r=\sqrt 5.
\end{cases}$$
Calculate the partial sum $S_9$ using the formula:
$$S_n=\dfrac{a_1(1-r^n)}{1-r}.$$
For $n=9$ we have:
$$\begin{align*}
S_{9}&=\dfrac{7\left(1-(\sqrt 5)^9\right)}{1-\sqrt 5}\\
&=\dfrac{7(1-625\sqrt 5)}{1-\sqrt 5}\\
&=\dfrac{7(1-625\sqrt 5)(1+\sqrt 5)}{(1-\sqrt 5)(1+\sqrt 5)}\\
&=\dfrac{7(-3124-624\sqrt 5}{-4}\\
&=7(781+156\sqrt 5).\end{align*}$$
