If in the expansion of (a-2b)^{n}, the sum of | vedclass

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(B) The general term in the expansion of $(a-2b)^{n}$ is given by $t_{r+1} = {}^{n}C_{r} (a)^{n-r} (-2b)^{r}$.Given that the sum of the $5^{th}$ and $

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

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(B) The general term in the expansion of $(a-2b)^{n}$ is given by $t_{r+1} = {}^{n}C_{r} (a)^{n-r} (-2b)^{r}$.Given that the sum of the $5^{th}$ and $6^{th}$ term is zero, we have $t_5 + t_6 = 0$.This implies ${}^{n}C_4 (a)^{n-4} (-2b)^4 + {}^{n}C_5 (a)^{n-5} (-2b)^5 = 0$.Rearranging the terms, we get ${}^{n}C_4 (a)^{n-4} (16b^4) = -{}^{n}C_5 (a)^{n-5} (-32b^5)$.Dividing both sides by ${}^{n}C_4 (a)^{n-5} (b^4)$, we get $a = -\frac{{}^{n}C_5}{{}^{n}C_4} (-2b)$.Using the formula ${}^{n}C_r = \frac{n!}{r!(n-r)!}$, we have $\frac{{}^{n}C_5}{{}^{n}C_4} = \frac{n-4}{5}$.Thus, $a = -\frac{n-4}{5} (-2b) = \frac{2(n-4)}{5} b$.Therefore, $\frac{a}{b} = \frac{2(n-4)}{5}$.

If in the expansion of $(a-2b)^{n}$, the sum of the $5^{th}$ and $6^{th}$ term is zero, then the value of $\frac{a}{b}$ is

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