If the coefficients of x^{7} and x^{8} in the | vedclass

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Question

${ }^{n} C_{7} 2^{n-7} \frac{1}{3^{7}}=^{n} C_{8} 2^{n-8} \frac{1}{3^{8}}$

$\Rightarrow \frac{n !}{(n-7) ! 7 !} 2^{n-7} \frac{1}{3^{7}}=\frac{n !}{

Vedclass · Accepted Answer

$55$

Vedclass · Answer

${ }^{n} C_{7} 2^{n-7} \frac{1}{3^{7}}=^{n} C_{8} 2^{n-8} \frac{1}{3^{8}}$

$\Rightarrow \frac{n !}{(n-7) ! 7 !} 2^{n-7} \frac{1}{3^{7}}=\frac{n !}{(n-8) ! 8 !} 2^{n-8} \frac{1}{3^{8}} \Rightarrow \frac{1}{(n-7)}=\frac{1}{8} \cdot \frac{1}{2} \cdot \frac{1}{3}$

$\Rightarrow n-7=48 \Rightarrow n=55$

If the coefficients of $x^{7}$ and $x^{8}$ in the expansion of $\left(2+\frac{x}{3}\right)^{n}$ are equal, then the value of $n$ is equal to $.....$

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