Four numbers are chosen at random (without replacement) from the set $$\{1, 2, 3, \ldots, 20\}$$. Statement-1: The probability that the chosen numbers when arranged in some order will form an AP is $$\frac{1}{85}$$. Statement-2: If the four chosen numbers form an AP, then the set of all possible values of common difference is $$\{\pm 1, \pm 2, \pm 3, \pm 4, \pm 5\}$$.

To test Statement-1 we must count how many unordered sets of four distinct numbers from $$\{1,2,\dots ,20\}$$ can be arranged to form an arithmetic progression (AP).

Write a four-term AP with first term $$x$$ and positive common difference $$d$$ as
$$x,\;x+d,\;x+2d,\;x+3d$$

Because every term must lie in $$\{1,\dots ,20\}$$ we need
$$1\le x \quad\text{and}\quad x+3d\le 20\;.$$ Hence $$x\le 20-3d$$, so for a fixed $$d$$ the admissible first term has the range
$$x=1,2,\dots ,20-3d\;.$$ The number of APs corresponding to one positive $$d$$ is therefore $$20-3d$$ (provided the right side is positive).

Case 1: Counting for each positive $$d$$

$$\begin{aligned} d=1&:;;20-3(1)=17\quad &\Longrightarrow&\;17\text{ APs}\\ d=2&:;;20-3(2)=14 &\Longrightarrow&\;14\text{ APs}\\ d=3&:;;20-3(3)=11 &\Longrightarrow&\;11\text{ APs}\\ d=4&:;;20-3(4)=8 &\Longrightarrow&\;8\text{ APs}\\ d=5&:;;20-3(5)=5 &\Longrightarrow&\;5\text{ APs}\\ d=6&:;;20-3(6)=2 &\Longrightarrow&\;2\text{ APs}\\ d\ge 7&:;;20-3d\le 0 &\Longrightarrow&\;0\text{ APs} \end{aligned}$$

The total number of distinct four-term APs is
$$17+14+11+8+5+2=57\;.$$ (Using negative $$d$$ would only reverse the same sets, so no new AP appears.)

Case 2: Computing the probability

The total number of unordered selections of four numbers from 20 is
$$\binom{20}{4}=4845\;.$$

Therefore the required probability is
$$\frac{57}{4845}=\frac{1}{85}\;.$$

Statement-1 is thus correct.

Case 3: Valid values of the common difference

From the list above, the possible positive differences are $$1,2,3,4,5,6$$, so the complete set of allowable differences is
$$\{\pm 1,\pm 2,\pm 3,\pm 4,\pm 5,\pm 6\}\;.$$ Statement-2 claims the set $$\{\pm 1,\pm 2,\pm 3,\pm 4,\pm 5\}$$, omitting $$\pm 6$$. Counter-examples such as $$\{1,7,13,19\}$$ (difference $$6$$) and $$\{2,8,14,20\}$$ (difference $$6$$) show that $$\lvert d\rvert=6$$ is indeed possible. Hence Statement-2 is false.

Statement-1 is true and Statement-2 is false.

Option B which is: Statement-1 is true, Statement-2 is false