10. For the sequence $\left\{a_{n}\right\}$, if there exists a sequence $\left\{b_{n}\right\}$, such that for any $n \in \mathbf{Z}_{+}$, we have $a_{n} \geqslant b_{n}$, then $\left\{b_{n}\right\}$ is called a "weak sequence" of $\left\{a_{n}\right\}$. Given
$$
\begin{array}{l}
a_{n}=n^{3}-n^{2}-2 t n+t^{2}\left(n \in \mathbf{Z}_{+}\right), \\
b_{n}=n^{3}-2 n^{2}-n+\frac{5}{4}\left(n \in \mathbf{Z}_{+}\right),
\end{array}
$$

and $\left\{b_{n}\right\}$ is a weak sequence of $\left\{a_{n}\right\}$. Then the range of the real number $t$ is
$\qquad$
10. $\left(-\infty, \frac{1}{2}\right] \cup\left[\frac{3}{2},+\infty\right)$.

From the problem, we know that for any positive integer $n$,
$$
f(n)=a_{n}-b_{n}=n^{2}-(2 t-1) n+t^{2}-\frac{5}{4} \geqslant 0.
$$

Then $\Delta=(2 t-1)^{2}-4\left(t^{2}-\frac{5}{4}\right)=-4 t+6$.
(1) When $\Delta \leqslant 0$, i.e., $t \geqslant \frac{3}{2}$, the condition holds.
(2) When $\Delta>0$, i.e., $t<\frac{3}{2}$, $\frac{2 t-1}{2}<1$.
Thus, the axis of symmetry of the quadratic function $f(n)$ is to the left of $n=1$. In this case, the necessary and sufficient condition for the condition to hold is
$$
\begin{array}{l}
f(1) \geqslant 0 \Leftrightarrow t^{2}-2 t+\frac{3}{4} \geqslant 0 \\
\Leftrightarrow t \leqslant \frac{1}{2} \text { or } t \geqslant \frac{3}{2} \Leftrightarrow t \leqslant \frac{1}{2}.
\end{array}
$$

From (1) and (2), we know that the range of $t$ is
$$
\left(-\infty, \frac{1}{2}\right] \cup\left[\frac{3}{2},+\infty\right).
$$