In the binomial expansion `(a-b)^n, nge5` the sum of 5th and 6th terms is zero. Then find `a/b`

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

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

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

If in the binomial expansion of (1+x)^(n) ,the coefficient of 14th,15th and 16th terms are in A.P.,then find n.

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)