在数列  中, 若 $a_{1}=1, a_{2}=\frac{1}{2}$,

$\frac{2}{a_{n+1}}=\frac{1}{a_{n}}+\frac{1}{a_{n+2}}\left(n \in N^{*}\right)$, 则该数列的通项为( ).
 $\because \frac{2}{a_{n+1}}=\frac{1}{a_{n}}+\frac{1}{a_{n+2}}\left(n \in N^{*}\right), \therefore$ 数列 $\left\{\frac{1}{a_{n}}\right\}$ 是等差数列,又 $\frac{1}{a_{2}}-\frac{1}{a_{1}}=2-1=1, \therefore \frac{1}{a_{n}}=1+(n-1)=n, \quad \therefore a_{n}=\frac{1}{n}$.

故选 A.