We can choose 3 tickets out of 21 tickets in C(21,3) ways.

 

The three-term A.P. is determined by its first and last elements, which must have the same parity.

 

 

There are 10 even numbers and 11 odd numbers out of 21.

 

So, if we count the ways to pick the highest and the lowest such that they have the same parity, then, the middle number can be selected in a specific way.

 

If the lowest is 1, then the middle number can be selected in 10 ways.

 

If the lowest is 2 or 3, then the middle number can be selected in 9 ways.

 

If the lowest is 4 or 5, then the middle number can be selected in 8 ways.

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If the lowest is 10, then the middle number can be selected in 1 ways.

 

So the total

=n(n+1)2+(n−1)n2=n2 

=n(n+1)2+(n−1)n2=n2 

 

 

Total number of arrangements = C(21,3)=1330

 

So, the probability of the tickets that are drawn in AP

=n(n+1)2 +(n−1)n2 =n2