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

T5=nC4an−4(−2b)4      General term =Tr+1 = ncr xn-r ar and T6=nC5an−5(−2b)5As T5+T6=0, we get  nC424an−4b4=nC525an−5b5or an−4b4an−5b5=n!255!(n−5)!⋅4!(n−4)!n!24or ab=2(n−4)5