4. In the arithmetic sequence $\left\{a_{n}\right\}$, if $a_{2}+a_{4}+a_{6}+$ $a_{8}+a_{10}=80$, then $a_{7}-\frac{1}{2} a_{8}=(\quad)$.
(A) 4
(B) 6
(C) 8
(D) 10
4.C.

Since $a_{2}+a_{4}+a_{6}+a_{8}+a_{10}=5 a_{6}=80$, therefore, $a_{6}=16$.
Thus, $a_{7}-\frac{1}{2} a_{8}=a_{6}+d-\frac{1}{2}\left(a_{6}+2 d\right)=\frac{1}{2} a_{6}=8$.