13.258. A train is traveling from station $A$ to station $B$. On a certain section of the route near station $B$, repair work was being carried out, and on this section, the train was allowed to travel at a speed that was only $1 / n$ of its original speed, as a result of which the train arrived at station $B$ with a delay of $a$ hours. The next day, the repair work front moved closer to station $B$ by $b$ km, and under the same conditions, the train was delayed by only $c$ hours. Find the speed of the train.
## Solution.

Let $x$ km/h be the speed of the train, $t$ h be the time the train takes to travel from $A$ to $B$ without stopping, and $S$ km be the distance the train travels after stopping.

According to the problem, 
\[
\left\{
\begin{array}{l}
\frac{x t - S}{x} + \frac{S}{\frac{x}{n}} = t + a, \\
\frac{x t - (S - b)}{x} + \frac{S - b}{\frac{x}{n}} = t + c,
\end{array}
\right.
\]
from which we get $x = \frac{b(n-1)}{a - c}, n > 1, a > c$.

Answer: $\frac{b(n-1)}{a - c}$ km/h, $n > 1, a > c$.