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 $$\frac{31}{40}$$ and $$a_3 + a_4 + a_5 = 14$$, then $$m + n$$ is equal to _____.

Let the five terms of the G.P. be

$$a,\ ar,\ ar^2,\ ar^3,\ ar^4$$

Given that their mean is $$\frac{31}{10}$$,

$$\frac{a+ar+ar^2+ar^3+ar^4}{5}=\frac{31}{10}$$

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

The mean of their reciprocals is $$\frac{31}{40}$$,

$$\frac{\frac1a+\frac1{ar}+\frac1{ar^2}+\frac1{ar^3}+\frac1{ar^4}}{5}=\frac{31}{40}$$

$$\frac{1+r+r^2+r^3+r^4}{ar^4}=\frac{31}{8}\tag{2}$$

Dividing (1) by (2),

$$a^2r^4=4$$

Since all terms are positive,

$$ar^2=2$$

Also,

$$ar^2+ar^3+ar^4=14$$

$$ar^2(1+r+r^2)=14$$

$$2(1+r+r^2)=14$$

$$1+r+r^2=7$$

$$r^2+r-6=0$$

$$(r+3)(r-2)=0$$

Since $$r>0$$,

$$r=2$$

Using $$ar^2=2$$,

$$a(2)^2=2$$

$$a=\frac12$$

Hence the five terms are

$$\frac12,\ 1,\ 2,\ 4,\ 8$$

Now,

$$\sum a_i^2=\left(\frac12\right)^2+1^2+2^2+4^2+8^2$$

$$=\frac14+1+4+16+64$$

$$=\frac{341}{4}$$

Variance,

$$\sigma^2=\frac{\sum a_i^2}{5}-(\text{Mean})^2$$

$$\sigma^2=\frac{\frac{341}{4}}{5}-\left(\frac{31}{10}\right)^2$$

$$=\frac{341}{20}-\frac{961}{100}$$

$$=\frac{1705-961}{100}$$

$$=\frac{744}{100}$$

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

Thus,

$$m=186,\quad n=25$$

$$m+n=186+25=211$$

$$\boxed{211}$$