If the coefficients of x^9 and x^{10} in the | vedclass

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(A) The general term in the expansion of $(3+\frac{x}{2})^n$ is given by $T_{r+1} = {}^nC_r (3)^{n-r} (\frac{x}{2})^r = {}^nC_r \frac{3^{n-r}}{2^r} x^

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$69$

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(A) The general term in the expansion of $(3+\frac{x}{2})^n$ is given by $T_{r+1} = {}^nC_r (3)^{n-r} (\frac{x}{2})^r = {}^nC_r \frac{3^{n-r}}{2^r} x^r$. The coefficient of $x^r$ is ${}^nC_r \frac{3^{n-r}}{2^r}$. For $r=9$,the coefficient is ${}^nC_9 \frac{3^{n-9}}{2^9}$. For $r=10$,the coefficient is ${}^nC_{10} \frac{3^{n-10}}{2^{10}}$. Given that these coefficients are equal: ${}^nC_9 \frac{3^{n-9}}{2^9} = {}^nC_{10} \frac{3^{n-10}}{2^{10}}$. Dividing both sides by ${}^nC_9 \frac{3^{n-10}}{2^{10}}$,we get: $\frac{3^{n-9}}{3^{n-10}} 	imes \frac{2^{10}}{2^9} = \frac{{}^nC_{10}}{{}^nC_9}$. $3^1 	imes 2^1 = \frac{n-10+1}{10} = \frac{n-9}{10}$. $6 = \frac{n-9}{10}$ $\Rightarrow n-9 = 60$ $\Rightarrow n = 69$. Thus,the correct option is $A$.

If the coefficients of $x^9$ and $x^{10}$ in the binomial expansion of $(3+\frac{x}{2})^n$ are equal,then $n=$

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