If the coefficient of the 3^{text{rd}} term f | vedclass

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(C) The general term in the expansion of $(x+y)^n$ is $T_{r+1} = ^nC_r x^{n-r} y^r$. For the expansion of $\left(ax^2 - \frac{8}{bx}\right)^9$,the $3^

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$a^5b^5 = -2$

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(C) The general term in the expansion of $(x+y)^n$ is $T_{r+1} = ^nC_r x^{n-r} y^r$. For the expansion of $\left(ax^2 - \frac{8}{bx}ight)^9$,the $3^{	ext{rd}}$ term from the beginning is $T_3$ $(r=2)$: $T_3 = ^9C_2 (ax^2)^{9-2} (-\frac{8}{bx})^2 = ^9C_2 a^7 x^{14} \cdot \frac{64}{b^2 x^2} = ^9C_2 \cdot \frac{64 a^7}{b^2} x^{12}$. The coefficient is $^9C_2 \cdot \frac{64 a^7}{b^2}$. For the expansion of $\left(ax - \frac{2}{bx^2}ight)^9$,the $3^{	ext{rd}}$ term from the end is the $3^{	ext{rd}}$ term from the beginning in the expansion of $\left(-\frac{2}{bx^2} + axight)^9$,which is $T_3$ $(r=2)$: $T_3 = ^9C_2 (-\frac{2}{bx^2})^{9-2} (ax)^2 = ^9C_2 (-\frac{2}{bx^2})^7 (ax)^2 = ^9C_2 \cdot (-\frac{128}{b^7 x^{14}}) \cdot a^2 x^2 = -^9C_2 \cdot \frac{128 a^2}{b^7 x^{12}}$. Wait,the question asks for the coefficient of the $3^{	ext{rd}}$ term from the end. In $(x+y)^n$,the $k^{	ext{th}}$ term from the end is the $(n-k+2)^{	ext{th}}$ term from the beginning. For $n=9, k=3$,this is the $(9-3+2) = 8^{	ext{th}}$ term from the beginning. $T_8 = ^9C_7 (ax)^{9-7} (-\frac{2}{bx^2})^7 = ^9C_7 (ax)^2 (-\frac{128}{b^7 x^{14}}) = ^9C_2 \cdot \frac{-128 a^2}{b^7 x^{12}}$. Equating the coefficients: $^9C_2 \cdot \frac{64 a^7}{b^2} = -^9C_2 \cdot \frac{128 a^2}{b^7}$. $\frac{a^7}{b^2} = -\frac{2 a^2}{b^7} \implies a^5 b^5 = -2$.

If the coefficient of the $3^{\text{rd}}$ term from the beginning in the expansion of $\left(ax^2 - \frac{8}{bx}\right)^9$ is equal to the coefficient of the $3^{\text{rd}}$ term from the end in the expansion of $\left(ax - \frac{2}{bx^2}\right)^9$,then the relation between $a$ and $b$ is:

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