6. Let $f(x)$ be a function defined on the set of real numbers $\mathbf{R}$, and satisfies the following relations:
$$
f(10+x)=f(10-x), f(20-x)=-f(20+x) \text {. }
$$

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
(C) $-1-2-1+1$
6. [Analysis and Solution] From the first given equation, we have
$$
\begin{array}{l}
f[10+(10-x)]=f[10-(10-x)] \\
\therefore f(x)=f(20-x)
\end{array}
$$

From the second given equation, we have
$$
\begin{array}{l}
f(x)=-f(20+x) \\
\therefore f(40+x)=f[20+(20+x)] \\
=-f(20+x)=f(x),
\end{array}
$$

which shows that $f(x)$ is a periodic function.
From (1) and (2), we get
$$
f(-x)=f(20+x)=-f(x) \text {. }
$$
$\therefore f(x)$ is an odd function, so the answer is $(\mathbf{C})$.