12. Let $t$ be a positive number, and the sequence $\left\{a_{n}\right\}$ is an arithmetic sequence with the first term $t$ and common difference $2 t$, and its sum of the first $n$ terms is $S_{n}$. If $\lim _{n \rightarrow \infty} \frac{\sqrt{S_{n}}}{a_{n}}>$ $\sqrt{1-t}$, then the range of values for $t$ is $\qquad$
12. $00)$, then
$$
\begin{array}{c}
a_{n}=t+(n-1) 2 t=(2 n-1) t, \\
S_{n}=n t+\frac{n(n-1)}{2} \cdot 2 t=n^{2} t .
\end{array}
$$

Thus, $\lim _{n \rightarrow \infty} \frac{\sqrt{S_{n}}}{a_{n}}=\lim _{n \rightarrow \infty} \frac{n \sqrt{t}}{(2 n-1) t}=\frac{1}{2 \sqrt{t}}$.
From $\frac{1}{2 \sqrt{t}}>\sqrt{1-t}$, we solve to get $0<t \leqslant 1$ and $t \neq \frac{1}{2}$.