6. In the sequence $\left\{a_{n}\right\}$, if the adjacent terms $a_{n} 、 a_{n+1}$ are the two roots of the quadratic equation
$$
x_{n}^{2}+3 n x_{n}+c_{n}=0(n=1,2, \cdots)
$$

then when $a_{1}=2$, $c_{100}=$ $\qquad$
6.22496.

From Vieta's formulas, we have
$$
\begin{array}{l}
a_{n}+a_{n+1}=-3 n, \\
a_{n} a_{n+1}=c_{n} .
\end{array}
$$

From equation (1), we get
$$
\begin{array}{l}
\left(a_{n+1}+a_{n+2}\right)-\left(a_{n}+a_{n+1}\right)=-3 \\
\Rightarrow a_{n+2}-a_{n}=-3 .
\end{array}
$$

Equation (3) indicates that $a_{1}, a_{3}, \cdots, a_{2 k-1}$ and $a_{2}, a_{4}, \cdots, a_{2 k}$ are both arithmetic sequences with a common difference of -3.

Given $a_{1}=2$, substituting into equation (1) yields $a_{2}=-5$. Thus, we obtain the arithmetic sequences:
$$
\begin{array}{l}
a_{2 k-1}=a_{1}+(k-1)(-3)=5-3 k, \\
a_{2 k}=a_{2}+(k-1)(-3)=-2-3 k .
\end{array}
$$

Therefore, $a_{100}=-2-3 \times 50=-152$,
$$
a_{101}=5-3 \times 51=-148 \text {. }
$$

Substituting into equation (2) gives
$$
c_{100}=a_{100} a_{101}=(-152)(-148)=22496 .
$$