Statement 1: Our of 5 tickets consecutively numbered, three are drawn at random. The chance that the numbers on them are in A.P. is 2/15. Statement 2: Out of `2n+1` tickets consecutively numbed, three are drawn at random, the chance that the numbers on them are in A.P. is `3n//(4n^2-1)dot`

Statement 1: Our of 5 tickets consecutively numbered, three are drawn at random. The chance that the numbers on them are in A.P. is 2//15 . Statement 2: Out of 2n+1 tickets consecutively numbered, three are drawn at random, the chance that the numbers on them are in A.P. is 3n//(4n^2-1) .

Statement 1: Out of 5 tickets consecutively numbered, three are drawn at random. The chance that the numbers on them are in A.P. is 2//15 . Statement 2: Out of 2n+1 tickets consecutively numbered, three are drawn at random, the chance that the numbers on them are in A.P. is 3n//(4n^2-1) .

Out of (2n+1) tickets consecutively numbered , three are drawn at random . The chance that the numbers on them are in A.P is

Out of (2n+1) tickets consecutively numbered, three are drawn at random. Find the chance that the numbers on them are in AP.

Out of (2n+1) tickets consecutively numbered, three are drawn at random. Find the chance that the numbers on them are in AP.

Out of (2n+1) tickets consecutively numbered, three are drawn at random. Find the chance that the numbers on them are in AP.

Out of (2n+1) tickets consecutively numbered, three are drawn at random. Find the chance that the numbers on them are in AP.

Out of (2n+1) tickets consecutively numbered, three are drawn at random. Find the chance that the numbers on them are in AP.

Statement-1: Out of 21 tickets with number 1 to 21, 3 tickets are drawn at random, the chance that the numbers on them are in AP is (10)/(133) . Statement-2: Out of (2n+1) tickets consecutively numbered three are drawn at ranodm, the chance that the number on them are in AP is (4n-10)/ (4n^(2)-1) .