Solution:
\( 5 \) terms of GP,
\(\displaystyle \frac{a}{r^2},\;\frac{a}{r},\;a,\;ar,\;ar^2 \)
Now, \(\displaystyle \frac{\displaystyle \frac{a}{r^2}+\frac{a}{r}+a+ar+ar^2}{5}=\frac{31}{10} \)
\(\displaystyle \dfrac{{a}\left\{\displaystyle\frac{1}{r^2}+\frac{1}{r}+1+r+r^2\right\}}{5}=\frac{31}{10}\quad \text{....(i)} \)
\( \displaystyle\dfrac{\left\{\displaystyle\frac{r^2}{a}+\frac{r}{a}+\frac{1}{a}+\frac{1}{ar}+\frac{1}{ar^2}\right\}}{5}=\frac{31}{40}\quad \text{....(ii)} \)
\(\dfrac{\dfrac{1}{a} \left\{ r^2 + r + 1 + \dfrac{1}{r} + \dfrac{1}{r^2} \right\}}{5} = \dfrac{31}{40}\text{....(ii)}\)
Now divide \(\text{eq}^n\) (i) from (ii)
\( \dfrac{a}{1/a}=\dfrac{31\times 40}{10\times 31}=4 \)
\( a^2=4\)
\(a=2 \) (as \( a \) is positive \( \Rightarrow a\neq -2 \))
Now, from \(\text{eq}^n\) (i)
\( \dfrac{{2}\left\{r^2+\displaystyle\frac{1}{r^2}+r+\frac{1}{r}+1\right\}}{5}=\dfrac{31}{10}\Rightarrow \left(r+\dfrac{1}{r}\right)^2-2\left(r+\dfrac{1}{r}\right)+1=\dfrac{31}{4} \)
Let \( r+\dfrac{1}{r}=\alpha \)
\( \alpha^2+\alpha-\dfrac{35}{4}=0 \)
\( 4\alpha^2+4\alpha-35=0 \)
\(4\alpha^2 + 14\alpha - 10\alpha - 35 = 0\)
\( 2\alpha(2\alpha+7)-5(2\alpha+7)=0 \)
\( \alpha={5}/{2},\;-{7}/{2}\quad\Rightarrow \alpha=\dfrac{5}{2}\;(\text{As } \alpha=r+\dfrac{1}{r}\Rightarrow \alpha>0) \)
\( r+\dfrac{1}{r}=\dfrac{5}{2} \)
\(2(r^2+1)=5r\)
\(\boxed{ r=2} \)
Now, numbers are,
\(\Rightarrow \dfrac{a}{r^2},\dfrac{a}{r},a,ar,ar^2 \quad\Rightarrow \dfrac{1}{2},1,2,4,8 \quad\Rightarrow a_3+a_4+a_5=14 \)
\(2+4+8=144\)
Now variance \( (\sigma^2) = \dfrac{\dfrac{1}{4} + 1 + 4 + 16 + 64}{5} - \left(\dfrac{31}{10}\right)^2 \)
\(\displaystyle=\frac{341}{20}-\frac{961}{100}=\frac{1705-961}{100}=\frac{744}{100}=\frac{186}{25} \)
\( m=186,\; n=25 \)
\(\boxed{ m+n=211} \)

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