If the 4^{text{th}} term in the expansion of | vedclass

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(B) The general term in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^nC_r a^{n-r} b^r$. For the expansion of $\left(\frac{x}{2}-\frac{2y}{3}\r

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$3$

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(B) The general term in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^nC_r a^{n-r} b^r$. For the expansion of $\left(\frac{x}{2}-\frac{2y}{3}ight)^6$,the $4^{	ext{th}}$ term $(T_4)$ corresponds to $r=3$. $T_4 = {}^6C_3 \left(\frac{x}{2}ight)^{6-3} \left(-\frac{2y}{3}ight)^3$. $T_4 = 20 	imes \left(\frac{x}{2}ight)^3 	imes \left(-\frac{8y^3}{27}ight)$. $T_4 = 20 	imes \frac{x^3}{8} 	imes \left(-\frac{8y^3}{27}ight) = -20 	imes \frac{x^3 y^3}{27}$. Given that $T_4 = -20$,we have $-20 	imes \frac{x^3 y^3}{27} = -20$. Dividing both sides by $-20$,we get $\frac{x^3 y^3}{27} = 1$. $x^3 y^3 = 27$. Taking the cube root of both sides,$xy = 3$.

If the $4^{\text{th}}$ term in the expansion of $\left(\frac{x}{2}-\frac{2y}{3}\right)^6$ is $-20$,then $xy=$

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