Binomial Theorem 1 Question 20
22. In the binomial expansion of $(a-b)^{n}, n \geq 5$ the sum of the 5 th and 6 th terms is zero. Then, $a / b$ equals
(a) $\frac{n-5}{6}$
(b) $\frac{n-4}{5}$
(c) $\frac{5}{n-4}$
(d) $\frac{6}{n-5}$
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Solution:
- Given, $T _5+T _6=0$
$\Rightarrow{ }^{n} C _4 a^{n-4} b^{4}-{ }^{n} C _5 a^{n-5} b^{5}=0$
$\Rightarrow{ }^{n} C _4 a^{n-4} b^{4}={ }^{n} C _5 a^{n-5} b^{5} \Rightarrow \frac{a}{b}=\frac{{ }^{n} C _5}{{ }^{n} C _4}=\frac{n-4}{5}$
