Show that $\mathbb{E}\left(\bar{X}_{n}\mid X_{(1)},X_{(n)}\right) = \frac{X_{(1)}+X_{(n)}}{2}$ 
Let $X_{1},\ldots,X_{n}$ be i.i.d. $U[\alpha,\beta]$ r.v.s., and let $X_{(1)}$ denote the $\min$, and $X_{(n)}$ the $\max$. Show that
  $$
\mathbb{E}\left(\overline{X}_{n}\mid X_{(1)},X_{(n)}\right) = \frac{X_{(1)}+X_{(n)}}{2}.
$$

I know that $\displaystyle\mathbb{E}\left(\overline{X}_{n}\mid X_{(1)},X_{(n)}\right)=\mathbb{E}\left({X}_{1}\mid X_{(1)},X_{(n)}\right)$ but not much more.

Assume (because of symmetry, it doesn't matter; and it simpifies notation) that $x_{(1)}=x_1=A$ and $x_{(n)}=x_n=B$ ; and $n\ge 3$. 
Then the variables $x_2, x_3 \cdots x_{n-1}$ are iid uniform on $[A,B]$
Hence $$E\left(\frac{\sum x_i}{n} \mid x_{(1)},x_{(2)}\right) =\frac{1}{n}\left(A+ (n-2)\frac{(A+B)}2+B \right)=\frac{A+B}{2}$$
