1. Given $a_{n}=\frac{n-\sqrt{98}}{n-\sqrt{99}}\left(n \in \mathbf{N}_{+}\right)$, among the first 20 terms of the sequence $\left\{a_{n}\right\}$, the maximum term and the minimum term are ( ).
A. $a_{1}, a_{20}$
B. $a_{1}, a_{9}$
C. $a_{10}, a_{9}$
D. $a_{10}, a_{20}$
1. C.

Since $a_{n}=1+\frac{\sqrt{99}-\sqrt{98}}{n-\sqrt{99}}$, when $n=10$ then $n-\sqrt{99}$ is the smallest positive number, so $a_{10}$ is the largest; when $n=9$ then $n-\sqrt{99}$ is the largest negative number, so $a_{9}$ is the smallest.