6.114 个.

Let $a_{i}(i=1,2,3,4) \in\{1,2,3, \cdots, 20\}$ and form an $AP$.
The common difference is $d$, then $d=\frac{a_{4}-a_{1}}{4-1}=\frac{a_{4}-a_{1}}{3} \in \mathbf{Z}$, so $a_{4}=a_{1}(\bmod 3)$. Therefore, we only need to classify $1,2, \cdots, 20$ by modulo 3.
Let $\left.P_{r}=\mid p H=3 q+r \quad(1 \leqslant p \leqslant 20) q, p \in \mathbf{Z} \quad 0 \leqslant r \leqslant 2\right\}$ Considering $20=3 \times 6+2$, so $\left\{P_{0}\right\}=6,\left|P_{1}\right|=\left|P_{2}\right|=7$.
Therefore, when $d>0$, there are a total of arithmetic sequences $C_{6}^{2}+2 C_{7}^{2}=57$; when $d<0$, there are another 57, so the answer is 114.
6. From the set $\{1,2,3, \cdots, 20\}$, choose any 4 different numbers such that these 4 numbers form an arithmetic sequence. There are $\qquad$ such sequences.