Q52 / 160 · WBJEE 2010 · Mathematics
If in the expansion of $(a-2 b)^{\mathrm{n}}$, the sum of the 5 th and 6 th term is zero, then the value of $\frac{\mathrm{a}}{\mathrm{b}}$ is
- A$\frac{\mathrm{n}-4}{5}$
- Correct answerB$\frac{2(n-4)}{5}$
- C$\frac{5}{n-4}$
- D$\frac{5}{2(n-4)}$
Solution
$$
\text { Hints: } \begin{gathered}
(a-2 b)^{\mathrm{n}}=\sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}(a)^{\mathrm{n}-\mathrm{r}}(-2 \mathrm{~b})^{\mathrm{r}} \\
\mathrm{t}_5+\mathrm{t}_6=0
\end{gathered}
$$
$$
\begin{aligned}
& \Rightarrow{ }^{\mathrm{n}} \mathrm{C}_4(\mathrm{a})^{\mathrm{n}-4}(-2 \mathrm{~b})^4+{ }^{\mathrm{n}} \mathrm{C}_5(\mathrm{a})^{\mathrm{n}-5}(-2 \mathrm{~b})^5=0 \Rightarrow \frac{\mathrm{n} !}{4 !(\mathrm{n}-4) !} \mathrm{a}^{\mathrm{n}-4}(-2 \mathrm{~b})^4=-\frac{\mathrm{n} !}{5 !(\mathrm{n}-5) !}(\mathrm{a})^{\mathrm{n}-5}(-2 \mathrm{~b})^5 \\
& \Rightarrow \frac{1}{(\mathrm{n}-4)} \times \mathrm{a}=\frac{-1}{5}(-2 \mathrm{~b}) \Rightarrow \frac{\mathrm{a}}{\mathrm{b}}=\frac{2(\mathrm{n}-4)}{5}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.