Let the 11 consecutive natural numbers be $1, 2, 3, \dots, 11.$ Total ways to choose any 3 numbers = $\displaystyle \binom{11}{3} = 165.$
Now, we need to count the number of 3-number selections that can form an arithmetic progression (A.P.) with positive common difference.
For an A.P., let the middle term be $a$ and common difference be $d>0$. Then the three terms are: $(a-d,\ a,\ a+d)$ These must all lie between $1$ and $11$.
That means $1 \le a-d$ and $a+d \le 11$ ⟹ $d \le \min(a-1,\ 11-a)$
Now we count possible values of $d$ for each $a$:

$a$	$\min(a-1,\ 11-a)$	Possible $d$ values
1	0	–
2	1	1
3	2	1,2
4	3	1,2,3
5	4	1,2,3,4
6	5	1,2,3,4,5
7	4	1,2,3,4
8	3	1,2,3
9	2	1,2
10	1	1
11	0	–

Total = $1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 25.$
Hence, number of favorable triplets = $25.$
Therefore, $\displaystyle P = \frac{25}{165} = \frac{5}{33}.$
Final Answer: $\boxed{\dfrac{5}{33}}$