11. (20 points) In the Cartesian coordinate system $x O y$, $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\frac{x^{2}}{2}+y^{2}=1$. Let the line $l$ intersect the ellipse at two distinct points $A$ and $B$, and suppose the line $l$ 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 $d$.
11. Given the conditions, we know the points $F_{1}(-1,0)$ and $F_{2}(1,0)$. Let the line $l: y=kx+m$, and the points $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$.
Then $x_{1}$ and $x_{2}$ satisfy $\frac{x^{2}}{2}+(kx+m)^{2}=1$, which is
$$
\left(2k^{2}+1\right)x^{2}+4kmx+\left(2m^{2}-2\right)=0.
$$

Since points $A$ and $B$ do not coincide and the slope of line $l$ exists, $x_{1}$ and $x_{2}$ are two distinct real roots of equation (1).
Therefore, the discriminant of equation (1) is
$$
\begin{array}{l}
\Delta=(4km)^{2}-4\left(2k^{2}+1\right)\left(2m^{2}-2\right) \\
=8\left(2k^{2}+1-m^{2}\right)>0 \\
\Rightarrow 2k^{2}+1>m^{2}.
\end{array}
$$

Given that the slopes of lines $A F_{1}$, $l$, and $B F_{1}$ are $\frac{y_{1}}{x_{1}+1}$, $k$, and $\frac{y_{2}}{x_{2}+1}$, respectively, and they form an arithmetic sequence, we have
$$
\begin{array}{l}
\frac{y_{1}}{x_{1}+1}+\frac{y_{2}}{x_{2}+1}=2k \\
\Rightarrow (kx_{1}+m)(x_{2}+1)+(kx_{2}+m)(x_{1}+1) \\
\quad=2k(x_{1}+1)(x_{2}+1) \\
\Rightarrow (m-k)(x_{1}+x_{2}+2)=0.
\end{array}
$$

Assume $m=k$. Then the equation of line $l$ is $y=kx+k$, which means $l$ passes through the point $F_{1}(-1,0)$, which does not meet the conditions.
Therefore, $x_{1}+x_{2}+2=0$.
Hence, by equation (1) and Vieta's formulas, we have
$$
\begin{array}{l}
\frac{4km}{2k^{2}+1}=-(x_{1}+x_{2})=2 \\
\Rightarrow m=k+\frac{1}{2k}.
\end{array}
$$

From equations (2) and (3), we have
$$
\begin{array}{l}
2k^{2}+1>m^{2}=\left(k+\frac{1}{2k}\right)^{2} \\
\Rightarrow k^{2}>\frac{1}{4k^{2}} \Leftrightarrow |k|>\frac{\sqrt{2}}{2}.
\end{array}
$$

Conversely, when $m$ and $k$ satisfy equation (3) and $|k|>\frac{\sqrt{2}}{2}$, line $l$ must not pass through point $F_{1}$ (otherwise, it would lead to $m=k$, contradicting equation (3)).

In this case, $m$ and $k$ satisfy equation (2), so line $l$ intersects the ellipse at two distinct points $A$ and $B$, and it ensures that the slopes of $A F_{1}$ and $B F_{1}$ exist (otherwise, one of $x_{1}$ or $x_{2}$ would be -1, and combining $x_{1}+x_{2}+2=0$, we would have $x_{1}=x_{2}=-1$, contradicting that equation (1) has two distinct real roots).
The distance from point $F_{2}(1,0)$ to line $l: y=kx+m$ is
$$
\begin{aligned}
d & =\frac{|k+m|}{\sqrt{1+k^{2}}}=\frac{1}{\sqrt{1+k^{2}}}\left|2k+\frac{1}{2k}\right| \\
& =\frac{1}{\sqrt{\frac{1}{k^{2}}+1}}\left(2+\frac{1}{2k^{2}}\right).
\end{aligned}
$$

Note that $|k|>\frac{\sqrt{2}}{2}$.
Let $t=\sqrt{\frac{1}{k^{2}}+1}$. Then $t \in(1, \sqrt{3})$.
Thus, equation (4) can be rewritten as
$$
d=\frac{1}{t}\left(\frac{t^{2}}{2}+\frac{3}{2}\right)=\frac{1}{2}\left(t+\frac{3}{t}\right).
$$

Considering that the function $f(t)=\frac{1}{2}\left(t+\frac{3}{t}\right)$ is monotonically decreasing on the interval $[1, \sqrt{3}]$, we have from equation (5)
$$
f(\sqrt{3})<d<f(1) \Rightarrow d \in(\sqrt{3}, 2).
$$