In the binomial expansion of `(a-b)^nngeq5,` the sum of the 5th and 6th term is zero. Then `a//b` equals `(n-5)//6` b. `(n-4)//5` c. `n//(n-4)` d. `6//(n-5)`

In the binomial expansion of (a-b)^n,ngeq5, the sum of the 5th and 6th term is zero. Then a//b equals (n-5)//6 b. (n-4)//5 c. n//(n-4) d. 6//(n-5)

In the binomial expansion of (a-b)^(n) , n ge 5 , the sum of 5th and 6th terms is zero, then (a)/(b) equals

In the binomial expansion of (a - b)^n , n ge 5 the sum of the 5th and 6th term is zero , then find a/b

In the binomial expansion of (a-b)^n , ngeq5, the sum of 5th and 6th terms is zero, then a/b equals (1) 5/(n-4) (2) 6/(n-5) (3) (n-5)/6 (4) (n-4)/5

If in the expansion of (a-2b)^n , the sum of 5th and 6th terms is 0, then the values of a/b a. (n-4)/5 b. (2(n-4))/5 c. 5/(n-4) d. 5/(2(n-4))

If in the expansion of (a-2b)^(n) , the sum of 5^(th) and 6^(th) terms is 0, then the values of a/b is (a) (n-4)/(5) (b) (2(n-4))/(5) (c) (5)/(n-4) (d) (5)/(2(n-4))

If the coefficient of 4th term in the expansion of (a+b)^n is 56, then n is

If the coefficient of 4th term in the expansion of (a+b)^n is 56, then n is

If in any binomial expansion a, b, c and d be the 6th, 7th, 8th and 9th terms respectively, prove that (b^2-ac)/(c^2-bd)=(4a)/(3c)

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