Let the positive numbers a _ 1 , a _ 2, a _ 3 , a _ 4 and a _ 5 be...

$5$ positive numbers ${a}_{1}, {a}_{2}, . . . , {a}_{5}$ are in geometric progression. So, let the numbers be $\dfrac{a}{{r}^{2}}, \dfrac{a}{r}, a, ar, a{r}^{2}$

According to question,

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

And,

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

$⇒ \dfrac{\dfrac{1}{a}\left({r}^{2} + r + 1 + \dfrac{1}{r} + \dfrac{1}{{r}^{2}}\right)}{5} = \dfrac{31}{40} . . . \left(2\right)$

By \left(1\right) \text{ and } \left(2\right)$, we get</p><p>${a}^{2} = 4 ⇒ a = 2$</p><p>So,</p><p>\left({r}^{2} + r + 1 + \dfrac{1}{r} + \dfrac{1}{{r}^{2}}\right) = \dfrac{31}{4}$

$⇒ {\left(r + \dfrac{1}{r}\right)}^{2} + r + \dfrac{1}{r} = \dfrac{27}{4} + 2$

$⇒ {t}^{2} + t = \dfrac{35}{4}$

where, $t = r + \dfrac{1}{r}$

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

$⇒ t = \dfrac{5}{2} ⇒ r = 2$

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

Variance is

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

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

$m + n = 211$

Hence this is the required option.