Let 2n + 1 numbers be
A-n , A-n+1 , A-n+2 ..... A-1 , A , A1 , A2 , A3 , A4 ...... An-1 , An ( Consecutive numbers )
Three numbers are in AP
Let say { a - r , r , a + r } where a is middle term of AP.
Now total ways to select 3 numbers out of (2n + 1 ) : C( 2n + 1 , 3 )
where C( n , r ) = binomial coefficient.
C(n,r) = (n/r) * C(n-1, r - 1) = (n/r) * ( (n-1)/(r-1) ) * ((n-2)/((r-2))....and so on till r != 1.
Using above binomial formula :
Total ways : ((2n + 1 ) * ( 2n ) * (2n-1))/6 = (n * (4n^2 -1) )/3
Now number of favourable cases.
Let say middle term a = A ( middle tern of (2n+1) numbers.
Then { a-r , a + r } will be { ( A-1 , A1 ) ( A-2 , A2) (A-3 , A3) .... (A-n+1 , An-1) , (A-n , An ) }
Total N ways.
Now if middle term r = A1 then we have (N-1) ways : { (A , A2 ) , ( A-1 , A3 ) .... and so on }
In this manner total ways : n + 2 * ( (n-1) + (n-2) + (n-3) .... 3 + 2 + 1 )
Middle term can be formed n triplets , next and previois term foemed ( n-1 ) triplets and so on...
Total favourable ways : n + 2 * (n*(n-1)/2 = n^2
Required Probality : n^2/ ((n * (4n^2 -1) )/3) : 3n/(4n^2-1)
Option (A) is the correct asnwer.