First slide

Binomial theorem for positive integral Index

Question

If in the expansion of $(a - 2 b)^{n}$ ,the sum of 4th and 5th term is zero, then the value of $\frac{a}{b}$ is

NA

A

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

$\frac{n - 3}{2}$
C

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

$\frac{5}{2 (n - 4)}$

Solution

Here, $T_{4} =^{n} C_{3} (a)^{n - 3} (- 2 b)^{3}$
and $T_{5} =^{n} C_{4} (a)^{n - 4} (- 2 b)^{4}$
Given $T_{4} + T_{5} = 0$
$\begin{aligned} \Rightarrow^{n} C_{3} (a)^{n - 3} (- 2 b)^{3} +^{n} C_{4} (a)^{n - 4} (- 2 b)^{4} = 0 \\ \Rightarrow (a)^{n - 4} (- 2 b)^{3} [a^{n} C_{3} +^{n} C_{4} (- 2 b)] = 0 \end{aligned}$
$\Rightarrow \frac{a}{b} = \frac{2^{n} C_{4}}{^{n} C_{3}}$
$= \frac{2 \cdot n (n - 1) (n - 2) (n - 3)}{4 \cdot 3 \cdot 2 \cdot 1} \times \frac{3 \cdot 2 \cdot 1}{n (n - 1) (n - 2)}$
$= \frac{n - 3}{2}$

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