A fair $n \left(n>1\right)$ 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

Given,

A fair $n \left(n>1\right)$ faces die is rolled repeatedly until a number less than $n$ appears,

$$\begin{gathered} x_i 1 2 3 4 ... \\ {p}_i \frac{n-1}{n} \frac{1}{n}·\left(\frac{n-1}{n}\right) \frac{1}{n^2}·\frac{n-1}{n} \frac{1}{n^3}·\left(\frac{n-1}{n}\right) ... \end{gathered}$$

Now we know that, mean is given by,

Mean $=\sum _{i=1}^{\infty }{p}_i{x}_i=1·\frac{n-1}{n}+\frac{2}{n}·\left(\frac{n-1}{n}\right)+\frac{3}{n^2}\left(\frac{ n-1}{n}\right)+...$

$\Rightarrow \frac{n}{9}=\left(1-\frac{1}{n}\right)S ........\left(1\right)$

Where, $S=1+\frac{2}{n}+\frac{3}{n^2}+\frac{4}{n^3}+... ......\left(2\right)$

$\frac{1}{n}S=\frac{1}{n}+\frac{2}{n^2}+\frac{3}{n^3}+... ......\left(3\right)$

Now subtracting equation $\left(2\right)-\left(3\right)$ we get,

$\left(1-\frac{1}{n}\right)S=1+\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+...$

$\Rightarrow \left(1-\frac{1}{n}\right)S=\frac{1}{1-{\displaystyle \frac{1}{n}}}$

Now putting the value of $S$ in equation $\left(1\right)$ we get,

$\Rightarrow \frac{n}{9}=\left(1-\frac{1}{n}\right)\times \frac{1}{{\left(1-{\displaystyle \frac{1}{n}}\right)}^2}=\frac{n}{n-1}$

$\Rightarrow n=10$