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

Then $f(x)$ is
(A) an even function and a periodic function;
(B) an even function, but not a periodic function;
(C) an odd function and a periodic function;
(D) an odd function, but not a periodic function.
6. (C)

From the first given equation, we get
$$
\begin{aligned}
& f[10+(10-x)]=f[10-(10-x)], \\
\therefore \quad & f(x)=f(20-x),
\end{aligned}
$$

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

It is evident that $f(x)$ is a periodic function.
From (1) and (2), we get
$$
f(-x)=f(20+x)=-f(x),
$$
$\therefore \quad f(x)$ is an odd function. Therefore, the answer is (C).