Since $a_1, a_2, a_3, a_4, a_5$ are in geometric progression, we can write :

$a_2 = r a$,

$a_3 = r^2 a$,

$a_4 = r^3 a$,

$a_5 = r^4 a$.

where $r$ is the common ratio and $a_1$ = $$a$$ is the first term.

Given that the mean of the series is $\frac{31}{10}$, we get

$$\frac{1}{5}(a_1 + a_2 + a_3 + a_4 + a_5) = \frac{31}{10}$$

Substituting the terms with the values of $a_1$ and $r$ gives

$$\frac{1}{5}(a + r a + r^2 a + r^3 a + r^4 a) = \frac{31}{10}$$

Simplifying this gives

$$a (1 + r + r^2 + r^3 + r^4) = \frac{31}{2}$$

$$ \begin{aligned} & \frac{a\left(r^5-1\right)}{r-1}=\frac{31}{2} ......(1) \\\\ & \frac{1}{a}\left(1+\frac{1}{r}+\frac{1}{r^2}+\frac{1}{r^3}+\frac{1}{r^4}\right)=\frac{31}{40} \cdot 5=\frac{31}{8} \\\\ & \frac{1}{a}\left(\frac{1-\left(\frac{1}{r}\right)^5}{1-\frac{1}{r}}\right)=\frac{31}{8} \\\\ & \text { or } \frac{1}{a}\left(\frac{r^5-1}{r-1}\right) \frac{1}{r^4}=\frac{31}{8} .........(2) \end{aligned} $$

From (1) and (2)

$$ \frac{1}{a} \cdot \frac{31}{2 a} \cdot \frac{1}{r^4}=\frac{31}{8} $$

$$ a r^2=2 $$

From (1)

$$ \begin{aligned} & \frac{2}{r^2}\left(\frac{r^5-1}{r-1}\right)=\frac{31}{2} \\\\ & \frac{1+r+r^2+r^3+r^4}{r^2}=\frac{31}{4} \\\\ & \left(r^2+\frac{1}{r^2}\right)+\left(r+\frac{1}{r}\right)=\frac{27}{4} \\\\ & t^2-2+t=\frac{27}{4} \end{aligned} $$

$$ \begin{aligned} &4 t^2+4 t-35=0\\\\ &4 t^2+14 t-10 t-35=0\\\\ &(2 t-5)(2 t+7)=0\\\\ &t=\frac{5}{2}, \frac{-7}{2} \Rightarrow r=2\\\\ & \therefore r=2, a=\frac{1}{2} \end{aligned} $$

$$ \text { Variance of data set }\left\{\frac{1}{2}, 1,2,4,8\right\} $$

$$ \therefore \sigma^2=\frac{\sum{\mathrm{X}^2}}{\mathrm{~N}}-\left(\frac{\sum{\mathrm{X}}}{\mathrm{N}}\right)^2 $$

$$ \begin{aligned} & =\frac{\left(\frac{341}{4}\right)}{5}-\left(\frac{31}{10}\right)^2 \\\\ & =\frac{341}{20}-\frac{961}{100}=\frac{1705-961}{100} \\\\ & =\frac{744}{100} \end{aligned} $$

$$ =\frac{186}{25}=\frac{\mathrm{m}}{\mathrm{n}} \Rightarrow 211=\mathrm{m}+\mathrm{n} $$