1.… a decreasing arithmetic sequence, the sum of the 5th powers of its first 7 terms is zero, and the sum of the 4th powers is 2002. Then the 7th term of this sequence is ( ).
(A) 2002
(B) -2002
(C) $3 \sqrt[4]{\frac{1001}{98}}$
(D) $-3 \sqrt[4]{\frac{1001}{98}}$
$-.1 .(\mathrm{b})$
Let the arithmetic sequence be: $a_{t}$ ! with a common difference of $d(d<0)$.
From $u_{1}^{5}+u_{2}^{5}+\cdots+u_{7}^{5}=0$, we get
$$
\begin{array}{l}
a_{4}=0, a_{5}=-a_{3}=d, a_{n}=-a_{2}=2 d, \\
a_{7}=-a_{1}=3 d .
\end{array}
$$

From $a_{1}^{4}+a_{2}^{4}+\cdots+a_{7}^{4}=2002$, we get
$$
2 d^{4}\left(1^{4}+2^{4}+3^{4}\right)=2002 \text {. }
$$
$\therefore d=-\sqrt[4]{\frac{1001}{98}}$. Therefore, $a_{7}=3 d=-3 \sqrt[4]{\frac{1001}{98}}$.