11 Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C: \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$, and points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ are on the ellipse $C$. If $x_{1}+x_{2}=\frac{1}{2}$, and $\overrightarrow{A F_{2}}=$ $\lambda \overrightarrow{F_{2} B}$, find the value of $\lambda$.
11 From $\overrightarrow{F_{2} A}=\lambda \overrightarrow{F_{2} B}$, we know that points $A$, $B$, and $F_{2}$ are collinear.
If line $A B \perp x$-axis, then $x_{1}=x_{2}=1$, which does not meet the requirement.
If line $A B$ has a slope, let it be $k$, then the equation of line $A B$ is $y=k(x-1)$.
From 
$$
\left\{\begin{array}{l}\frac{x^{2}}{4}+\frac{y^{2}}{3}=1, \\ y=k(x-1),\end{array}\right.
$$ 
we get
$$
\left(4 k^{2}+3\right) x^{2}-8 k^{2} x+4 k^{2}-12=0 .
$$

Since the discriminant of (1) is
$$
\Delta=64 k^{4}-4\left(4 k^{2}+3\right)\left(4 k^{2}-12\right)=144\left(k^{2}+1\right)>0,
$$

we have $x_{1}+x_{2}=\frac{8 k^{2}}{4 k^{2}+3}, x_{1} x_{2}=\frac{4 k^{2}-12}{4 k^{2}+3}$.
From $x_{1}+x_{2}=\frac{8 k^{2}}{4 k^{2}+3}=\frac{1}{2}$, we get $k^{2}=\frac{1}{4}$.
Substituting $k^{2}=\frac{1}{4}$ into equation (1), we get
$$
4 x^{2}-2 x-11=0,
$$

solving for $x$ gives $x=\frac{1 \pm 3 \sqrt{5}}{4}$.
$$
\text { Also, } \overrightarrow{A F_{2}}=\left(1-x_{1},-y_{1}\right), \overrightarrow{F_{2} B}=\left(x_{2}-1, y_{2}\right), \overrightarrow{F_{2} A}=\lambda \overrightarrow{F_{2} B} \text {, }
$$
$\lambda=\frac{1-x_{1}}{x_{2}-1}$, so $\lambda=\frac{3-\sqrt{5}}{2}$ or $\lambda=\frac{3+\sqrt{5}}{2}$.