Question

In the binomial expansion of $ {{(a-b)}^{n}},\,\,n\,\,\ge \,5, $ the sum of $5^{th}$ and $6^{th}$ term is zero. Then, $ \frac{a}{b} $ is equal to

• A. $ \frac{n-4}{2} $
• B. $ \frac{n-4}{3} $
• C. $ \frac{n-4}{5} $
• D. $ \frac{n-4}{4} $

Answer 1

Answer: (C)

Given expansion is $ {{(a-b)}^{n}}. $
$ \therefore $ $ {{T}_{5}}{{=}^{n}}{{C}_{4}}{{(a)}^{(n-4)}}{{(-b)}^{4}} $
and $ {{T}_{6}}{{=}^{n}}{{C}_{5}}{{(a)}^{(n-5)}}{{(-b)}^{5}} $
According to the given condition,
$ {{T}_{5}}+{{T}_{6}}=0 $
$ \therefore $ $ ^{n}{{C}_{4}}{{(a)}^{n-4}}{{(-b)}^{4}}{{+}^{n}}{{C}_{5}}{{(a)}^{n-5}}{{(-b)}^{5}}=0 $
$ \Rightarrow $ $ ({{a}^{n-4}}){{(-b)}^{4}}\left[ ^{n}{{C}_{4}}{{+}^{n}}{{C}_{5}}\left( \frac{-b}{a} \right) \right]=0 $
$ \Rightarrow $ $ \frac{a}{b}=\frac{^{n}{{C}_{5}}}{^{n}{{C}_{4}}}=\frac{\frac{n(n-1)\,(n-2)\,(n-3)\,(n-4)}{5\times 4\times 3\times 2\times 1}}{\frac{n(n-1)\,(n-2)\,(n-3)}{4\times 3\times 2\times 1}} $
$ =\frac{(n-4)}{5} $