10. If the points $A\left(x_{1}, y_{1}\right)$, $B\left(4, \frac{9}{5}\right)$, and $C\left(x_{2}, y_{2}\right)$ on the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$ are such that their distances to the right focus of the ellipse form an arithmetic sequence, and the perpendicular bisector $l$ of the line segment $A C$ intersects the $x$-axis at point $T$, find the equation of the line $B T$.
10. Let $a$, $b$, and $c$ represent the lengths of the semi-major axis, semi-minor axis, and semi-focal distance of an ellipse, respectively. Then
$$
a=5, b=3, c=4 \text {, }
$$

The right focus $F(4,0)$, and the equation of the directrix is $x=\frac{a^{2}}{c}$.
From $\frac{A F}{\frac{a^{2}}{c}-x_{1}}=\frac{c}{a}, \frac{C F}{\frac{a^{2}}{c}-x_{2}}=\frac{c}{a}$
$$
\Rightarrow A F=5-\frac{4}{5} x_{1}, C F=5-\frac{4}{5} x_{2} \text {. }
$$

According to the problem, we know that $A F+C F=2 B F$.
And $B F=\frac{9}{5}$, so
$$
\begin{array}{l}
\left(5-\frac{4}{5} x_{1}\right)+\left(5-\frac{4}{5} x_{2}\right)=\frac{18}{5} \\
\Rightarrow x_{1}+x_{2}=8 .
\end{array}
$$

Let the midpoint of segment $A C$ be $D$. Then $D\left(4, \frac{y_{1}+y_{2}}{2}\right)$. Suppose point $T\left(x_{0}, 0\right)$. Then the equation of the perpendicular bisector $D T$ of $A C$ is
$$
y-\frac{y_{1}+y_{2}}{2}=-\frac{x_{1}-x_{2}}{y_{1}-y_{2}}(x-4) .
$$

Since point $T\left(x_{0}, 0\right)$ lies on line $D T$, we have
$$
\begin{aligned}
& 0-\frac{y_{1}+y_{2}}{2}=-\frac{x_{1}-x_{2}}{y_{1}-y_{2}}\left(x_{0}-4\right) \\
\Rightarrow & x_{0}-4=\frac{y_{1}^{2}-y_{2}^{2}}{2\left(x_{1}-x_{2}\right)} .
\end{aligned}
$$

According to the points $A$ and $B$ on the ellipse, we get
$$
\begin{array}{l}
y_{1}^{2}=\frac{9}{25}\left(25-x_{1}^{2}\right), y_{2}^{2}=\frac{9}{25}\left(25-x_{2}^{2}\right) \\
\Rightarrow y_{1}^{2}-y_{2}^{2}=-\frac{9}{25}\left(x_{1}+x_{2}\right)\left(x_{1}-x_{2}\right) .
\end{array}
$$

From equation (1), we have
$$
\frac{y_{1}^{2}-y_{2}^{2}}{2\left(x_{1}-x_{2}\right)}=-\frac{36}{25} \text {. }
$$

Combining equations (2) and (3), we get $x_{0}=\frac{64}{25}$, i.e., point $T\left(\frac{64}{25}, 0\right)$.
Thus, the equation of line $B T$ is $25 x-20 y=64$.