AP or No AP?

Replace $10$ by $d$ and $11$ by $d+1$ . If $d+1$ is prime, the answer is $n = d^2+d+1$ .

Proof: Here is the key fact: any arithmetic progression of length $d+1$ must either contain an element divisible by $d+1$ or have difference divisible by $d+1$ . (Note that this is only true because $d+1$ is prime! Proof left as an exercise.)

So consider the subset of $\{ 1, 2, \ldots, n \}$ obtained by removing the multiples of $d+1$ , and suppose that $n < d^2+d+1$ . Then there can't be an AP of length $d+1$ whose difference is divisible by $d+1$ , since the difference between its first and last term would be at least $d^2+d$ . And we've picked this subset so that it has no multiples of $d+1$ in it. So it cannot have an AP of length $d+1$ . Also, there are at most $d$ multiples of $d+1$ in $\{ 1,2,\ldots, n \}$ . So the conclusion is that if $n < d^2+d+1$ , we can remove the mutiples of $d+1$ (plus whatever other numbers you want to get the total up to $d$ ) and avoid all APs of length $d+1$ .

On the other hand, if $n = d^2+d+1$ , and we want to avoid all APs of length $d+1$ , then we must remove at least one of the elements congruent to $1$ mod $d+1$ , in order to get rid of the AP $1, d+2, \ldots, d^2+d+1$ . Suppose we remove $k(d+1)+1$ . Then we have to remove at least $k$ elements less than this to avoid an AP of difference $1$ in the subset, and we have to remove at least $d-k$ elements greater than this in order to avoid an AP of difference $1$ from among the remaining $(d-k)(d+1)$ bigger elements; but that is a total of at least $d+1$ elements we'd have to remove, which is not allowed. So there is always an AP of length $d+1$ if $n = d^2+d+1$ ; hence this is the answer.

for ap of length 11 among natural numbers,let take first 110 numbers ,so 11,22 upto 110 total 10 numbers removed and if we just add 1 ,an ap of length 11 is formed

I don' like the solution but the solution is somewhat like this: Split the set from 1 to $n$ in parts of $11$ and then remove elements which are distinct modulo $11$ from each set. For $n$ to be least consider the common difference to be the least i.e $1$ so that the number of elements in the set is maximum(Though not required); So from $\Big\{1,2,...,11\Big\}$ remove (say 1) and then from $\Big\{12,13,...,23\Big\}$ remove 12(say) and so on and finally remove 110 $\Big\{99,,...,110\Big\}$ (Look at the right shift of the index of the number you remove from each set). Since 10 numbers have been already removed and there is no more AP in the set $\Big\{1,2,...,110\Big\}$ (Prove it). Adding one number to the set further makes an AP. To be conclusive, since we have removed one element each of type $i \mod 11$ with distinct $i$ each time so strictly speaking, now we are left with only 10 elements of each type in the set $\Big\{1,2,...,110\Big\}$ . Just add one number and we've formed an A.P.