6. Let $f(x)$ be a function defined on the set of real numbers $\mathbf{R}$, and satisfies the following relations: $f(10+x)=f(10-x)$, $f(20-x)=-f(20+x)$. Then $f(x)$ is
A. an even function, and also a periodic function
B. an even function, but not a periodic function
C. an odd function, and also a periodic function
D. an odd function, but not a periodic function
$$
\begin{array}{l}
f(20+x)=f[10+(10+x)]=f[10-(10+x)]=f(-x), \\
f(20-x)=f[10+(10-x)]=f[10-(10-x)]=f(x) \Rightarrow f(-x)=-f(x) \\
\Rightarrow f(20+x)=-f(x) \Rightarrow f(40+x)=-f(20+x)=f(x)
\end{array}
$$
$\Rightarrow f(x)$ is an odd function, and also a periodic function, so the answer is $C$.