Suppose that $M > \mu > m \geq 0$ are all fixed integers and assume that ${\bf p} \in \mathcal{P}(\mathbb N)$ is a probability distribution with support on $\{m,m+1,\ldots,M\}$ given by $$p_n = 0 ~~\text{for all $n \notin \{m,m+1,\ldots,M\}$}, ~~~\text{and}~~ p_n = p_m\cdot r^{n-m} ~~\text{for all $m\leq n \leq M$}$$ where $r \in (0,1) \cup (1,\infty)$ is the parameter (which plays the role of the "common ratio" in the usual geometric distribution). Assume also that the geometric-like distribution ${\bf p}$ defined above has a mean value equal to $\mu$ (i.e, $\sum_{m\leq n\leq M} np_n = \mu$ and $\sum_{m\leq n\leq M} p_n = 1$). I am wondering if it is possible to have a explicit formula which allows us to determine the unknown $r$ (and hence $p_m$ as a by-product) from $\mu$, $m$ and $M$ (which are all pre-fixed) ? It seems that there is no standard name for such a distribution in literature and the closest name I have found so far is the "geometric distribution on a bounded domain".
Remark: The question is equivalent to the following problem: Given integers $M > \mu > m \geq 0$ and suppose that $$\mu \sum_{m\leq n\leq M} r^n = \sum_{m\leq n\leq M} nr^n \tag{1}\label{1}.$$ Is it possible to solve for $r \in (0,1)\cup (1,\infty)$ from \eqref{1} which gives us a formula for $r$ in terms of $m$, $\mu$ and $M$? Notice that in the special case where $m = 0$ and $M \to \infty$, it is quite easy to obtain $r = \frac{\mu}{1+\mu}$.
Edit: I realized that $r$ can be larger than $1$ if $\mu$ is very close to $M$ and be relative large compared to $m$. Say for instance, $m = 2$, $\mu = 9$ and $M = 10$.
In order to find $r$ and $p_m$ basically we'll use both facts you mentioned. First we'll need to normalize the probability mass function: $$\sum_{n=m}^Mp_n=\sum_{n=m}^Mp_mr^{n-m}=p_m\sum_{n=0}^{M-m}r^n=p_m\frac{1-r^{M-m+1}}{1-r}\overset{!}{=}1\iff p_m=\frac{1-r}{1-r^{M-m+1}}\ (\ast)$$ At last, we'll find the closed form of its mean value in terms of the fixed and wanted parameters: $$ \begin{aligned} \sum_{n=m}^Mnp_n=p_m\sum_{n=m}^Mnr^{n-m}&=p_m\sum_{n=0}^{M-m}(n+m)r^n\\ &=p_m\sum_{n=0}^{M-m}nr^n+mp_m\sum_{n=0}^{M-m}r^n\\ &=rp_m\frac{\partial}{\partial r}\left(\sum_{n=0}^{M-m}r^n=\frac{1-r^{M-m+1}}{1-r}\right)+m\\ &=rp_m\left(\frac{1-r^{M-m+1}}{(1-r)^2}-\frac{(M-m+1)r^{M-m}}{1-r}\right)+m\\ &\overset{(\ast)}{=}r\left(\frac{1}{1-r}-\frac{(M-m+1)r^{M-m}}{1-r^{M-m+1}}\right)+m\\ &=\frac{r}{1-r}+(M-m+1)\frac{1-r^{M-m+1}-1}{1-r^{M-m+1}}+m\\ &=\frac{r}{1-r}-\frac{M-m+1}{1-r^{M-m+1}}+M+1\overset{!}{=}\mu\\ \end{aligned} $$ $$\iff (M-m+2)r-r^{M-m+2}+m-M-1=(\mu-M-1)(1-r)(1-r^{M-m+1})\\ \iff (M-\mu)r^{M-m+2}+(\mu-M-1)r^{M-m+1}+(\mu-m+1)r+m-\mu=0$$ Usually you'd solve this later equation computationally. Maybe you could've solved for $r$ in terms of $p_m$ in the first equation and do the same process, but still, the problem wouldn't become more simple. At last, notice it now also satisfies the limit you told me, namely, that when $M\to\infty$ and $m\to 0$ we get $$r=\frac{\mu}{1+\mu}.$$