2020/010) If the binomial expansion of (a-b)n, n≥5, the sum of the 5th and 6th terms is zero.

The $\frac{a}{b}$ equals  a. $\frac{n-5}{6}$ b. $\frac{n-4}{5}$ c. $\frac{5}{n-4}$ d. $\frac{6}{n-5}$ 

Solution:
we have

$5^{th}$ term = ${n \choose 4} a^{n-4} (-b)^4$
$6^{th}$ term = ${n \choose 5} a^{n-5} (-b)^5$

As sum of the two is zero we have ${n \choose 4} a^{n-4} (-b)^4 + {n \choose 5} a^{n-5} (-b)^5 = 0$

or ${n \choose 4} a - {n \choose 5} (b) = 0$

or $\frac{a}{b} = \frac{n \choose 5}{n \choose 4}$

$= \frac{\frac{n!}{5!(n-5)!}}{\frac{n!}{4!(n-4)!}}$

$= \frac{(n-4)!}{(n-5)! 5}= \frac{n-4}{5}$

Hence ans (b)