1. Total number of ways to choose 4 numbers from $\{1,2,3,\dots,20\}$:

2. Condition for 4 chosen numbers to form an AP

If 4 distinct numbers can be arranged in AP, then they must be of the form

with $d\neq 0$ and all terms lying in $\{1,2,\dots,20\}$.

Since order of selection does not matter, we only count increasing APs with $d>0$.

3. Possible values of common difference

For the largest term to be at most 20,

Since $a\ge 1$,

So possible positive integer values are

Hence Statement-2, which says the possible values are $\left(\pm1,\pm2,\pm3,\pm4,\pm5\right)$, is false because $d=\pm 6$ is also possible. For example,

is an AP with common difference $6$.

4. Count the number of 4-term APs

For each fixed positive $d$, the first term $a$ can be chosen such that

So number of APs for each $d$ is $20-3d$.

Now compute:

• For $d=1$: $20-3=17$
• For $d=2$: $20-6=14$
• For $d=3$: $20-9=11$
• For $d=4$: $20-12=8$
• For $d=5$: $20-15=5$
• For $d=6$: $20-18=2$

Total favorable selections:

5. Required probability

Simplify:

So Statement-1 is true.

6. Conclusion about the statements

• Statement-1: True
• Statement-2: False

Therefore the correct option is:

7. Comparison with stored correct answer

Stored correct answer = B.

My derived answer also = B. Hence they agree.