5.  For the circle $x^{2}+y^{2}-5 x=0$, passing through the point $\left(\frac{5}{2}, \frac{3}{2}\right)$, if $d \in\left(\frac{1}{6}, \frac{1}{3}\right]$, then the set of values for $n$ is ( ).
(A) $\{4,5,6\}$
(B) $\{6,7,8,9\}$
(C) $\{3,4,5\}$
(D) $\{3,4,5,6\}$
5. (D).

It is easy to know that the longest chord passing through the point $\left(\frac{5}{2}, \frac{3}{2}\right)$ is the diameter, whose length is 5, and the shortest chord perpendicular to this diameter has a length of 4. Let the lengths of $n$ chords forming an arithmetic sequence be $a_{1}, a_{2}, \cdots, a_{n}$. Then $d=$ $\frac{a_{n}-a_{1}}{n-1}$.
$\because a_{1} \geqslant 4, a_{n} \leqslant 5$,
$\therefore d \leqslant \frac{1}{n-1}$.
If $d \in\left(\frac{1}{6}, \frac{1}{3}\right]$, then $\frac{1}{6}<\frac{1}{n-1} \leqslant \frac{1}{3}$.
Solving this, we get $n=4,5,6$.
Furthermore, if the sum of the terms of the arithmetic sequence of chords is not necessarily $4 \frac{11}{13}$, then $d=\frac{4}{13}\left(\frac{1}{6},-\frac{1}{3}\right]$, which also meets the requirements, in this case, there are only 3 chords. Therefore, the possible values of $n$ are $\{3,4,5,6\}$.