5. Given the set $A_{n}=\left\{x \mid 2^{n}<x<2^{n+1}\right.$, and $x=7 m$ $+1, n, m \in \mathbf{N}\}$. Then the sum of all elements in $A_{6}$ is ( ).
(A) 792
(B) 890
(C) 891
(D) 990
5. C.

Since $64=2^{6}<x<2^{7}=128$, and $x=7m+1$, thus $a_{1}$ $=71, a_{n}=127$. Given the common difference $d=7$, we find $n=9$. Therefore, $S_{y}=$ $\frac{a_{1}+a_{9}}{2} \times 9=891$.