Suppose $a_0=2017,a_1{a}_2,\dots ,a_{n-1},2023$$=a_n$ an are in $A.P$. Let $S=\frac{1}{2^{n+1}}\sum _{r=0}^n \left({ }^n{C}_r\right)a_r-1000$

Then $S$ is equal to 

 

Put ${ }^n{C}_r=C_r$ and let 

$T=a_0{C}_0+a_1{C}_1+a_2{C}_2+\dots +a_n{C}_n$ 

Using, $C_r=C_{n-r}{C}_r=C_{n-r}$

$T=a_n{C}_0+a_{n-1}{C}_1+\dots +a_0{C}_n$

Adding $\left(1\right)$ and $\left(2\right)$, we get

$2T=\left(a_0+a_n\right)C_0+\left(a_1+a_{n-1}\right)C_1+\left(a_2+a_{n-2}\right)C_2+\dots +\left(a_n+a_0\right)C_n$

But $a_0+a_n=a_1+a_{n-1}=a_2+a_{n-2}=\dots =a_n+a_0=2017+2023$

$\begin{array}{r}\therefore 2T=4040\left(C_0+C_1+\dots +C_n\right)\\ \Rightarrow T=\left(2020\right)2^n=\left(1010\right)2^{n+1}\end{array}$

 Thus, $S=\frac{1}{2^{n+1}}\left(T\right)-1000=10$