if `6n` tickets numbered 0,1,2,....... 6n-1 are placed in a bag and three are drawn out , show that the chance that the sum of the numbers on then is equal to 6n is `(3n)/((6n-1)(6n-2))`

if 6n tickets numbered 0126n-1 are placed in a bag and three are drawn out,show that the chance that the sum of the numbers on them is equal to 6n is (3n)/((6n-1)(6n-2))

Find the sum of the series whose nth term is given by: (4n^(3)+6n^(2)+2n)

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

Find the sum of n terms of the series whose nth term is 12n^2-6n+5 .

Find the sum of n terms of the series whose n^(th) term is 18 n^(2) - 6n + 5

Statement 1: Our of 5tickets 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) .

Find the sum of n terms of an AP where a_(n)=5-6n and n in N

Find the sum of n terms of the sequence (a_(n)), where a_(n)=5-6n,n in N

Solve : (2n + 3)/(6n -5) = 1

Find the sum to n terms of an A.P.whose nth terms is t_(n)=5+6n,n in N