Three, it is known that a decreasing arithmetic sequence, the sum of the fifth powers of its first 7 terms is 0, the sum of the fourth powers of its first 7 terms is 2000, find the seventh term of this sequence.
Let the first seven terms be $a-3d, a-2d, a-d, a, a+d, a+2d, a+3d$, then the given conditions are
$$
\begin{array}{l}
a\left(7a^{4}+20a^{2}d^{2} \times 14+10 \times 98d^{4}\right)=0 \\
7a^{4}+12a^{2}d^{2} \times 14+2 \times 98d^{4}=1960
\end{array}
$$

From (2), we know that the expression inside the parentheses in (1) is positive, so $a=0$. Substituting $a=0$ into (2) gives
$$
d^{4}=10
$$

Since the sequence is decreasing, $d=\sqrt[4]{10}$.
$$
a_{7}=3 \sqrt[4]{10} .
$$