77 Ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$ on: 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 focus $F(4,0)$ have distances that form an arithmetic sequence. If the perpendicular bisector of line segment $A C$ intersects the $x$-axis at point $T$, then the slope $k$ of line $B T$ is $\qquad$.
$77 \frac{5}{4}$. The equation of the ellipse is
$$
\frac{x^{2}}{25}+\frac{y^{2}}{9}=1
$$

The foci are $F(4,0)$, and the directrix is $x=\frac{25}{4}$, so we have
$$
|A F|=5-\frac{4}{5} x_{1}, \quad|C F|=5-\frac{4}{5} x_{2}
$$

Given that
$$
2|B F|=|A F|+|C F|,
$$

we have
$$
x_{1}+x_{2}=8
$$
The equation of the perpendicular bisector of $A C$ is
$$
y-\frac{y_{1}+y_{2}}{2}=-\frac{x_{1}-x_{2}}{y_{1}-y_{2}}\left(x-\frac{x_{1}+x_{2}}{2}\right),
$$

Taking $y=0$ gives
$$
X_{T}=\frac{y_{1}^{2}-y_{2}^{2}}{2\left(x_{1}-x_{2}\right)}+\frac{x_{1}+x_{2}}{2} .
$$

Since $A$ and $B$ are on the ellipse, we have
$$
\begin{array}{l}
9 x_{1}^{2}+25 y_{1}^{2}=225 \\
9 x_{2}^{2}+25 y_{2}^{2}=225
\end{array}
$$
(1) - (2) gives
$$
\frac{y_{1}^{2}-y_{2}^{2}}{x_{1}-x_{2}}=-\frac{9}{25}\left(x_{1}+x_{2}\right),
$$

Thus,
$$
X_{T}=-\frac{9}{25} \cdot \frac{x_{1}+x_{2}}{2}+\frac{x_{1}+x_{2}}{2}=\frac{64}{25}
$$

Therefore,
$$
k_{B T}=\frac{\frac{9}{5}}{4-\frac{64}{25}}=\frac{5}{4}
$$