A fair $n(n1)$ faces die is rolled repeatedly until a number less than $n$ appears. If the mean of

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A fair $n(n>1)$ faces die is rolled repeatedly until a number less than $n$ appears. If the mean of the number of tosses required is $\frac{n}{9}$, then $n$ is equal to

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(10) : Mean $=1 \cdot \frac{n-1}{n}+2 \frac{1}{n}\left(\frac{n-1}{n}ight)+3\left(\frac{1}{n}ight)^{2}\left(\frac{n-1}{n}ight)+\ldots$\ $	herefore \quad \frac{n}{9}=\left(\frac{n-1}{n}ight)\left(1+2\left(\frac{1}{n}ight)+3\left(\frac{1}{n}ight)^{2}+\ldotsight)$\ $\Rightarrow \quad \frac{n}{9}=\left(\frac{n-1}{n}ight)\left(1-\frac{1}{n}ight)^{-2}=\left(\frac{n-1}{n}ight) \cdot \frac{n^{2}}{(n-1)^{2}}$\ $\Rightarrow \quad \frac{n}{9}=\frac{n}{n-1} \Rightarrow n=10$\