If p and q are real numbers such that the 7^{ | vedclass

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(B) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^nC_r a^{n-r} b^r$. For the $7^{	ext{th}}$ term,$r = 6$. $T_7 =

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$9$

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(B) The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^nC_r a^{n-r} b^r$. For the $7^{	ext{th}}$ term,$r = 6$. $T_7 = {}^8C_6 \left(\frac{5}{p^3}ight)^{8-6} \left(-\frac{3q}{7}ight)^6 = 700$. ${}^8C_6 = \frac{8 	imes 7}{2} = 28$. $28 	imes \left(\frac{5}{p^3}ight)^2 	imes \left(\frac{3q}{7}ight)^6 = 700$. $28 	imes \frac{25}{p^6} 	imes \frac{3^6 q^6}{7^6} = 700$. $\frac{700}{28 	imes 25} = \frac{q^6}{p^6} 	imes \frac{3^6}{7^6}$. $1 = \frac{q^6}{p^6} 	imes \frac{3^6}{7^6} \Rightarrow \frac{q^6}{p^6} = \frac{7^6}{3^6}$. Taking the $6^{	ext{th}}$ root,$\frac{q}{p} = \frac{7}{3}$. $3q = 7p \Rightarrow 9q^2 = 49p^2$.

If $p$ and $q$ are real numbers such that the $7^{\text{th}}$ term in the expansion of $\left(\frac{5}{p^3} - \frac{3q}{7}\right)^8$ is $700$,then $49p^2 =$ (in $q^2$)

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