11. In the Cartesian coordinate system $x O y$, $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\frac{x^{2}}{2}+y^{2}=1$. Suppose a line $l$ intersects the ellipse at two distinct points $A$ and $B$ and does not pass through the focus $F_{1}$. The distance from the focus $F_{2}$ to the line $l$ is $d$. If the slopes of the lines $A F_{1}$, $l$, and $B F_{1}$ form an arithmetic sequence, find the range of values for $d$.
Let the line $l: x=k y+m(m \neq-1)$, and combine the equations
$$
\left\{\begin{array}{l}
x^{2}+2 y^{2}=2, \\
x=k y+m
\end{array} \Rightarrow\left(k^{2}+2\right) y^{2}+2 k m y+m^{2}-2=0 .\right.
$$

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), F_{1}(-1,0)$, then
$$
\begin{array}{l}
y_{1}+y_{2}=-\frac{2 k m}{k^{2}+2}, y_{1} y_{2}=\frac{m^{2}-2}{k^{2}+2} . \\
\text { and } \Delta=4 k^{2} m^{2}-4\left(k^{2}+2\right)\left(m^{2}-2\right)=8\left(k^{2}-m^{2}+2\right)>0 .
\end{array}
$$

Thus, $k_{A F_{1}}+k_{B F_{1}}=\frac{y_{1}}{x_{1}+1}+\frac{y_{2}}{x_{2}+1}=\frac{y_{1}\left(k y_{2}+m+1\right)+y_{2}\left(k y_{1}+m+1\right)}{\left(k y_{1}+m+1\right)\left(k y_{2}+m+1\right)}$
$$
=\frac{2 k y_{1} y_{2}+(m+1)\left(y_{1}+y_{2}\right)}{k^{2} y_{1} y_{2}+k(m+1)\left(y_{1}+y_{2}\right)+(m+1)^{2}}
$$
$=\frac{2 k\left(m^{2}-2\right)-2 k m(m+1)}{k^{2}\left(m^{2}-2\right)-2 k^{2} m(m+1)+(m+1)^{2}\left(k^{2}+2\right)}=\frac{-4 k-2 k m}{2(m+1)^{2}-k^{2}}=\frac{2}{k}$
$\Rightarrow 2(m+1)^{2}+k^{2}(m+1)=0 \Rightarrow 2(m+1)+k^{2}=0 \Rightarrow m=-\frac{k^{2}}{2}-1$,
Substitute into $\Delta>0 \Rightarrow 4 k^{2}-\left(k^{2}+2\right)^{2}+8>0 \Rightarrow 0<k^{2}<2$.
Thus, $d=\frac{\left|\frac{k^{2}}{2}+2\right|}{\sqrt{k^{2}+1}}=\frac{k^{2}+4}{2 \sqrt{k^{2}+1}}=\frac{1}{2}\left(\sqrt{k^{2}+1}+\frac{3}{\sqrt{k^{2}+1}}\right)$,
Note that the above function is monotonically decreasing on $(0,2)$, so $d \in(\sqrt{3}, 2)$.