已知等差数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和为 $S_{n}, a_{5}=5, S_{5}=15$, 则数列 $\left\{\frac{1}{a_{n} a_{n+1}}\right\}$ 的前 100 项和为 $(\quad)$
设等差数列的公差为 $\mathrm{d}$,由题意可得, $\left\{\begin{array}{c}a_{1}+41=5 \ 5 a_{1}+10 d=15\end{array}\right.$解方程可得, $d=1, a_{1}=1$ 由等差数列的通项公式可得, $a_{n}=a_{1}+(n-1) d=1+(n-1) \times 1=n \quad \therefore$$$\frac{1}{a_{n} a_{n+1}}=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} \quad S_{100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\ldots+\frac{1}{100}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}$$