13.4 .5 * The ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$ has three different points $A\left(x_{1}, y_{1}\right), B\left(4, \frac{9}{5}\right), C\left(x_{2}, y_{2}\right)$, and the distances from these points to the focus $F(4,0)$ form an arithmetic sequence. If the perpendicular bisector of segment $A C$ intersects the $x$-axis at point $T$, find the slope $k$ of line $B T$.
Given that $a=5, e=\frac{4}{5}$. As shown in the figure, it is easy to know that $|A F|=5-\frac{4}{5} x_{1},|C F|=5-$ $\frac{4}{5} x_{2}$. Since $2|B F|=|A F|+|C F|$, we have $\left(5-\frac{4}{5} x_{1}\right)+\left(5-\frac{4}{5} x_{2}\right)=\frac{18}{5}$. Simplifying, we get $x_{1}+$ $x_{2}=8$. (1) The midpoint of line segment $A C$ is $(4$, $\left.\frac{y_{1}+y_{2}}{2}\right)$, and the equation of its perpendicular bisector is $y-\frac{y_{1}+y_{2}}{2}$ $=-\frac{x_{1}-x_{2}}{y_{1}-y_{2}}(x-4)$.
(2) Since $A$ and $C$ are on the ellipse, we have $9 x_{1}^{2}+25 y_{1}^{2}=25 \times 9,9 x_{2}^{2}+$ $25 y_{2}^{2}=25 \times 9$. Subtracting the two equations, we get $9\left(x_{1}^{2}-x_{2}^{2}\right)+25\left(y_{1}^{2}-y_{2}^{2}\right)=0$.
(3) Substituting (1) into (3), simplifying, we get $-\frac{x_{1}-x_{2}}{y_{1}-y_{2}}=\frac{25\left(y_{1}+y_{2}\right)}{72}$.
(4) Substituting (4) into (2), and setting $y=0$, we get $4-x=\frac{36}{25}$, where $x$ is the x-coordinate of point $T$. Therefore, $k_{B T}=\frac{\frac{9}{5}-0}{4-x}=\frac{5}{4}$.