10. (This sub-question is worth 15 points) If 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 right focus of the ellipse form an arithmetic sequence, the perpendicular bisector $l$ of line segment $A C$ intersects the $X$-axis at point $T$. Find the equation of the line $B T$.
Let $a, b, c$ represent the semi-major axis, semi-minor axis, and semi-focal distance of the ellipse, respectively, with $a=5, b=3, c=4$. The right focus is $F(4,0)$, and the equation of the directrix is $x=\frac{25}{4}$. From
$$
\frac{A F}{\frac{25}{4}-x_{1}}=\frac{c}{a}, \quad \frac{C F}{\frac{25}{4}-x_{2}}=\frac{4}{5}
$$

we get
$$
A F=5-\frac{4}{5} x_{1}, C F=5-\frac{4}{5} x_{2}
$$

According to the arithmetic mean, $A F+C F=2 B F$, and $B F=\frac{9}{5}$, i.e., $\left(5-\frac{4}{5} x_{1}\right)+\left(5-\frac{4}{5} x_{2}\right)=\frac{18}{5}$, so $x_{1}+x_{2}=8$.
(1) (5 points)
Let the midpoint of segment $A C$ be $D$, then its coordinates are $D\left(4, \frac{y_{1}+y_{2}}{2}\right)$. Suppose the coordinates of point $T$ are $T\left(x_{0}, 0\right)$, 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 $T\left(x_{0}, 0\right)$ lies on this line, we have $0-\frac{y_{1}+y_{2}}{2}=-\frac{x_{1}-x_{2}}{y_{1}-y_{2}}\left(x_{0}-4\right)$, i.e., $x_{0}-4=\frac{y_{1}^{2}-y_{2}^{2}}{2\left(x_{1}-x_{2}\right)}$.

Also, since $A, B$ are on the ellipse, we get $y_{1}^{2}=\frac{9}{25}\left(25-x_{1}^{2}\right), y_{2}^{2}=\frac{9}{25}\left(25-x_{2}^{2}\right)$, so
$$
y_{1}^{2}-y_{2}^{2}=-\frac{9}{25}\left(x_{1}+x_{2}\right)\left(x_{1}-x_{2}\right)
$$

From (1), (2), and (3), we get $x_{0}=\frac{64}{25}$, i.e., the coordinates of point $T$ are $T\left(\frac{64}{25}, 0\right)$, thus the equation of line $B T$ is $25 x-20 y=64$.