12. (2018 浙江高三专题练习) 设数列 $\left\{a_{n}\right\}$ 满足 $a_{1}=1$, 且 $a_{n+1}-a_{n}=n+1\left(n \in N^{*}\right)$, 则数列 $\left\{a_{n}\right\}$ 的通项公式为 __ $\left\{\frac{1}{a_{n}}\right\} 前 10 项的和为
$\because a_{1}=1, a_{n+1}-a_{n}=n+1, \therefore a_{2}-a_{1}=2, a_{3}-a_{2}=3, \ldots, a_{n}-a_{n-1}=n(n \geq 2)$, 将以上 $n-1$个式子相加得 $a_{n}-a_{1}=2+3+\ldots+n=\frac{(n+2)(n-1)}{2}$, 即 $a_{n}=\frac{n(n+1)}{2}$,令 $b_{n}=\frac{1}{a_{n}}$, 则 $b_{n}=\frac{2}{n(n+1)}=2\left(\frac{1}{n}-\frac{1}{n+1}\right)$,故 $S_{10}=b_{1}+b_{2}+\ldots+b_{10}=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3} \cdots \cdots+\frac{1}{10}-\frac{1}{11}\right)=\frac{20}{11}$.