Question

No. of ways 3 tickets can be drawn from ( (2 n+1) ) tickets ( =(2 n+1) C(3) )
Say, ( mathrm{E}= ) event that 3 cards drawn will have nos. in ( mathrm{AP} )
There will be ( 2 n- 1 ) groups of 3 numbers each which are in ( mathrm{AP} )
When common difference ( =1, ) groups are ( {(1,2,3), ) ( (2,3,4), ldots,(2 n-1,2 n, 2 n+1)} )
When c.d. ( =2, ) groups are ( {(1,3,5),(2,4,6), ldots,(2 n-3,2 n-1, ) ( 2 n+1)} )
When c.d. =n, groups are ( {(1, n+1,2 n+1)} )
So, total no. of group of 3 nos. drawn from ( (2 n+1) ) which are in ( mathrm{AP} )
( =(2 n-1)+(2 n-3)+(2 n-5)+ldots+3+1 )
( =mathrm{n}^{wedge} 2 )
So, probability of the same ( =n^{wedge} 2 /(2 n+1) C(3)= ) ( 3 n /left(4 n^{wedge} 2-1right) )

# W.E.28:- From (2n+1) consecutive positive integers 3 are selected at random. The 3n probability that they are in A.P.=72

Solution