3. From the set $\{1,2,3, \cdots, 20\}$, any 4 different numbers are taken such that these 4 numbers form an arithmetic sequence. The number of such arithmetic sequences is
A. 110
B. 114
C. 116
D. 120
3. B For an arithmetic sequence with a common difference of $d$ (assuming $d>0$) and the first term $a$: $a, a+d, a+2d, a+3d$, since these four numbers are all in the set $\{1,2, \cdots, 20\}$, we have $1 \leqslant a \leqslant 20$ and $a+3d \leqslant 20$, thus $1 \leqslant a \leqslant 20-3d$, so $20-3d \geqslant 1$, which means $d \leqslant 6$ (since $d \in \mathbf{N}^{+}$).

Therefore, the number of sequences that satisfy the condition is $\sum_{d=1}^{6}(20-3d)=57$. Since $a+3d, a+2d, a+d, a$ also satisfy the condition, there are a total of $57 \times 2=114$ sequences that meet the condition.