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If the 6 th term in $\left(\frac{2 p}{3}+\frac{3 q}{2}\right)^9$ is " $a p^b q^{c \prime}$, then $a, b$ and $c$ respectively are
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Verified Answer
The correct answer is:
$189,4,5$
The sixth term in the expansion $\left(\frac{2 p}{3}+\frac{3 q}{2}\right)^9$ is
$$
\begin{aligned}
T_6 & =T_{5+1}={ }^9 C_5\left(\frac{2 p}{3}\right)^4\left(\frac{3 q}{2}\right)^5 \\
& =\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2} \times\left(\frac{2}{3}\right)^4\left(\frac{3}{2}\right)^5 p^4 q^5 \\
& =189 p^4 q^5=a p^b q^c
\end{aligned}
$$
(given)
On comparing, we get
$$
a=189, b=4 \text { and } c=5
$$
$$
\begin{aligned}
T_6 & =T_{5+1}={ }^9 C_5\left(\frac{2 p}{3}\right)^4\left(\frac{3 q}{2}\right)^5 \\
& =\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2} \times\left(\frac{2}{3}\right)^4\left(\frac{3}{2}\right)^5 p^4 q^5 \\
& =189 p^4 q^5=a p^b q^c
\end{aligned}
$$
(given)
On comparing, we get
$$
a=189, b=4 \text { and } c=5
$$
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