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,
$T_5+T_6=0$
$\therefore$ ${}^n{C}_4{(a)}^{n-4}{(-b)}^4{+}^n{C}_5{(a)}^{n-5}{(-b)}^5=0$
$\Rightarrow$ $(a^{n-4}){(-b)}^4\left[{}^n{C}_4{+}^n{C}_5\left(\frac{-b}{a}\right)\right]=0$
$\Rightarrow$ $\frac{a}{b}=\frac{{}^n{C}_5}{{}^n{C}_4}=\frac{\frac{n(n-1)(n-2)(n-3)(n-4)}{5\times 4\times 3\times 2\times 1}}{\frac{n(n-1)(n-2)(n-3)}{4\times 3\times 2\times 1}}$
$=\frac{(n-4)}{5}$