Let $X_1, X_2, \ldots , X_n$ be a random sample from a Uniform$(0, θ)$ population with $θ > 0$, where $n$ is an even integer. Let $M=\frac{X_{\left(\frac{n}{2}\right)}+X_{\left(\frac{n}{2} + 1\right)}}{2}$ be the median of the random sample. Find the joint pdf of $X_{(n/2)}$ and $M$.
Here I find that $$f_{X_{(n/2)}} = \frac{n!}{\left(\frac{n-2}{n}\right)!\left(\frac{n}{2}\right)!}\left(\frac{x}{\theta}\right)^{\frac{n-2}{2}}\frac{1}{\theta}\left(1-\frac{x}{\theta}\right)^{n/2}.$$
But I not sure how to find the rest. Can I have some guidance?
First note that you can set $n=2k$, $k=1,2,3,...$
Using the Order Statistics you have that the joint density $(X_{(k)};X_{(k+1)})$ is
$$f_{X_{(k)},X_{(k+1)}}(x,y)=\frac{(2k)!}{[\theta(k-1)!]^2}\left(\frac{x}{\theta}\right)^{k-1}\cdot \left(1-\frac{y}{\theta}\right)^{k-1}\cdot\mathbb{1}_{(x;\theta)}(y)$$
Given this joint density, you are requested to calculate the joint density of
$$\left( X_{(k)}; \frac{X_{(k)}+X_{(k+1)}}{2}\right)$$
I think you can proceed by yourself using standard method of transformation to solve the problem (i.e. the jacobian method)
To use the jacobian, simply set (remember that $0<x<y<\theta$
$$\begin{cases} z=\frac{x+y}{2} \\ u=x \end{cases}$$
easy calculate the jacobian that is $|J|=2$ and then just substituting in the joint density you get
$$f_{UZ}(u,z)=\frac{2(2k)!}{[\theta(k-1)!]^2}\left(\frac{u}{\theta}\right)^{k-1}\cdot \left(1-\frac{2z-u}{\theta}\right)^{k-1}\cdot\mathbb{1}_{(?;?)}(u)\cdot\mathbb{1}_{(?;?)}(z)$$