3. Given the sequences $\left\{a_{n}\right\},\left\{b_{n}\right\}$ are both arithmetic sequences, the sequence $\left\{c_{n}\right\}$ satisfies $c_{n}=a_{n} b_{n}\left(n \in \mathbf{N}^{*}\right)$. If $c_{1}=1440, c_{2}=1716, c_{3}=1848$, then $c_{8}=$ $\qquad$ .
3. 348 .

It is known that $c_{n}$ is a quadratic trinomial, and let $d_{n}=c_{n+1}$ (where $n$ is a non-negative integer), $d_{n}=a n^{2}+b n+c$. Substituting $n=0,1,2$ respectively, we get
$$
\left\{\begin{array}{l}
c=1440, \\
a+b+c=1716, \\
4 a+2 b+c=1848 .
\end{array}\right.
$$

Solving these, we get $a=-72, b=348, c=1440$.
$$
\text { Therefore, } c_{8}=d_{7}=-72 \times 7^{2}+348 \times 7+1440=348 \text {. }
$$