(12) The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{2}=\frac{1}{2}$, and $a_{n}\left(a_{n-1}+a_{n+1}\right)=2 a_{n+1}$ $a_{n-1}(n \geqslant 2)$, then the 2012th term of the sequence $\left\{a_{n}\right\}$ is ( ).
(A) $\frac{1}{2010}$
(B) $\frac{1}{2011}$
(C) $\frac{1}{2012}$
(D) $\frac{1}{2013}$
$12 \mathrm{C}$ hint:
$$
a_{n}\left(a_{n-1}+a_{n+1}\right)=2 a_{n+1} a_{n-1} \Rightarrow \frac{1}{a_{n+1}}+\frac{1}{a_{n-1}}=2 \frac{1}{a_{n}}(n \geqslant 2) .
$$

Therefore, $\left\{\frac{1}{a_{n}}\right\}$ is an arithmetic sequence, and $\frac{1}{a_{1}}=1, d=1$, so the general term formula of the sequence is $a_{n}=\frac{1}{n}$, hence the 2012th term is $\frac{1}{2012}$.