14. Given an ellipse centered at the origin $O$, with foci on the $x$-axis, and an eccentricity of $\frac{\sqrt{3}}{2}$, the ellipse passes through the point $\left(\sqrt{2}, \frac{\sqrt{2}}{2}\right)$. Let a line $l$ that does not pass through the origin $O$ intersect the ellipse at points $P$ and $Q$, and the slopes of the lines $OP$, $PQ$, and $OQ$ form a geometric progression. Find the range of the area of $\triangle OPQ$.
14. From the problem, we can set the equation of the ellipse as
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0) \text {. }
$$

Notice,
$$
\left\{\begin{array} { l } 
{ \frac { c } { a } = \frac { \sqrt { 3 } } { 2 } , } \\
{ \frac { 2 } { a ^ { 2 } } + \frac { 1 } { 2 b ^ { 2 } } = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=2, \\
b=1 .
\end{array}\right.\right.
$$

Therefore, the equation of the ellipse is $\frac{x^{2}}{4}+y^{2}=1$.
From the problem, we know that the slope of line $l$ exists and is not 0. Hence, we set the equation of line $l$ as
$$
y=k x+m(m \neq 0) .
$$

Notice, points $P\left(x_{1}, y_{1}\right)$ and $Q\left(x_{2}, y_{2}\right)$ satisfy
$$
\left\{\begin{array}{l}
y=k x+m, \\
x^{2}+4 y^{2}-4=0 .
\end{array}\right.
$$

Eliminating $y$ gives
$$
\begin{array}{l}
\left(1+4 k^{2}\right) x^{2}+8 k m x+4\left(m^{2}-1\right)=0 . \\
\text { Then } \Delta=64 k^{2} m^{2}-16\left(1+4 k^{2}\right)\left(m^{2}-1\right) \\
=16\left(4 k^{2}-m^{2}+1\right)>0, \\
x_{1}+x_{2}=-\frac{8 k m}{1+4 k^{2}}, x_{1} x_{2}=\frac{4\left(m^{2}-1\right)}{1+4 k^{2}} . \\
\text { Hence } y_{1} y_{2}=\left(k x_{1}+m\right)\left(k x_{2}+m\right) \\
=k^{2} x_{1} x_{2}+k m\left(x_{1}+x_{2}\right)+m^{2} .
\end{array}
$$

Since the slopes of lines $O P$, $P Q$, and $O Q$ form a geometric sequence, we have
$$
\begin{array}{l}
\frac{y_{1}}{x_{1}} \cdot \frac{y_{2}}{x_{2}}=\frac{k^{2} x_{1} x_{2}+k m\left(x_{1}+x_{2}\right)+m^{2}}{x_{1} x_{2}}=k^{2} \\
\Rightarrow-\frac{8 k^{2} m^{2}}{1+4 k^{2}}+m^{2}=0 .
\end{array}
$$

Since $m \neq 0$, we have $k^{2}=\frac{1}{4} \Rightarrow k= \pm \frac{1}{2}$.
Since the slope of line $O Q$ exists and is not 0, and $\Delta>0$ $\Rightarrow 0<m^{2}<2$, and $m^{2} \neq 1$.
Let $d$ be the distance from point $O$ to line $l$. Then
$$
\begin{array}{l}
S_{\triangle O P Q}=\frac{1}{2} d|P Q| \\
=\frac{1}{2} \cdot \frac{|m|}{\sqrt{1+k^{2}}} \cdot \sqrt{1+k^{2}}\left|x_{1}-x_{2}\right| \\
=\frac{1}{2}|m| \sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\
=\sqrt{m^{2}\left(2-m^{2}\right)} .
\end{array}
$$

Therefore, the range of $S_{\triangle O P Q}$ is $(0,1)$.