If the 6^{th} term in left(frac{2p}{3} + frac | vedclass

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

Vedclass · Accepted Answer

$189, 4, 5$

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(B) The general term in the expansion of $(x+y)^n$ is given by $T_{r+1} = {}^nC_r x^{n-r} y^r$. For the expansion $\left(\frac{2p}{3} + \frac{3q}{2}ight)^9$,the $6^{th}$ term $(T_6)$ is $T_{5+1}$. Here,$n=9$,$r=5$,$x=\frac{2p}{3}$,and $y=\frac{3q}{2}$. $T_6 = {}^9C_5 \left(\frac{2p}{3}ight)^{9-5} \left(\frac{3q}{2}ight)^5$ $T_6 = {}^9C_5 \left(\frac{2p}{3}ight)^4 \left(\frac{3q}{2}ight)^5$ $T_6 = \frac{9 	imes 8 	imes 7 	imes 6}{4 	imes 3 	imes 2 	imes 1} 	imes \left(\frac{2^4 p^4}{3^4}ight) 	imes \left(\frac{3^5 q^5}{2^5}ight)$ $T_6 = 126 	imes \frac{16 p^4}{81} 	imes \frac{243 q^5}{32}$ $T_6 = 126 	imes \frac{1}{81} 	imes 243 	imes \frac{16}{32} p^4 q^5$ $T_6 = 126 	imes 3 	imes \frac{1}{2} p^4 q^5 = 189 p^4 q^5$. Comparing $189 p^4 q^5$ with $ap^bq^c$,we get $a=189, b=4, c=5$.

If the $6^{th}$ term in $\left(\frac{2p}{3} + \frac{3q}{2}\right)^9$ is $ap^bq^c$,then $a, b$ and $c$ respectively are

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