Let the positive numbers $a_1, a_{2,} a_3, a_4$ and $a_5$ be in a G.P. Let their mean and variance be $\frac{31}{10}$ and $\frac{m}{n}$ respectively, where $m$ and $n$ are co-prime. If the mean of their reciprocals is $\frac{31}{10}$ and $a_3+a_4+a_5=14$, then $m+n$ is equal to ____________.

$5$ positive numbers $a_1, a_2,..., a_5$ are in geometric progression. So, let the numbers be $\frac{a}{r^2}, \frac{a}{r}, a, ar, a{r}^2$

According to question,

$\frac{{\displaystyle \frac{a}{r^2}+\frac{a}{r}+a+ar+a{r}^2}}{5}=\frac{31}{10} ...\left(1\right)$

And,

$\frac{{\displaystyle \frac{r^2}{a}+\frac{r}{a}+\frac{1}{a}+\frac{1}{ar}+\frac{1}{a{r}^2}}}{5}=\frac{31}{40}$

$\Rightarrow \frac{{\displaystyle \frac{1}{a}\left(r^2+r+1+\frac{{\displaystyle 1}}{{\displaystyle r}}+\frac{{\displaystyle 1}}{{\displaystyle r^2}}\right)}}{5}=\frac{31}{40} ...\left(2\right)$

By $\left(1\right) \& \left(2\right)$, we get

$a^2=4 \Rightarrow a=2$

So,

$\left(r^2+r+1+\frac{{\displaystyle 1}}{{\displaystyle r}}+\frac{{\displaystyle 1}}{{\displaystyle r^2}}\right)=\frac{31}{4}$

$\Rightarrow {\left(r+\frac{1}{r}\right)}^2+r+\frac{{\displaystyle 1}}{{\displaystyle r}}=\frac{27}{4}+2$

$\Rightarrow t^2+t=\frac{35}{4}$

where, $t=r+\frac{1}{r}$

$4t^2+4t-35=0$

$\Rightarrow t=\frac{5}{2}\Rightarrow r=2$

So, numbers are $\frac{1}{2}, 1, 2, 4, 8$

Variance is

$=\frac{{\displaystyle \frac{1}{4}}+1+4+16+64}{5}-{\left(\frac{{\displaystyle \frac{1}{2}}+1+2+4+8}{5}\right)}^2$

$=\frac{186}{25}=\frac{m}{n}$

$m+n=211$

Hence this is the required option.