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Q.

In the binomial expansion of ${(a - b)}^{n}$ , n>5 the sum of $5^{t h} a n d 6^{t h}$ terms is zero, then $\frac{a}{b}$ equals

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a

$\frac{n - 4}{5}$

b

$\frac{5}{n - 4}$

c

$\frac{6}{n - 5}$

d

$\frac{n - 5}{6}$

answer is A.

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Detailed Solution

${(a - b)}^{n}$

Given that $T_{5} + T_{6} = 0$

$\begin{array}{rcl} {}^{n}{c_{4}} {(a)}^{n - 4} {(- b)}^{4} & = & {}^{- n}{C_{5}} {(a)}^{n - 5} {(- b)}^{5} \\ {}^{n}{C_{4}} a & = & {}^{n}{C_{5}} b \\ \frac{a}{b} & = & \frac{{}^{n}{C_{5}}}{{}^{n}{C_{4}}} = \frac{n - 5 + 1}{5} = \frac{n - 4}{5} \end{array}$

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In the binomial expansion of (a-b)n, n>5 the sum of 5th and 6th terms is zero, then ab equals