Let $b_1 = b$, $b_2 = bq$, $b_3 = bq^2$, $b_4 = bq^3$, $b_5 = bq^4$ denote elements of the progression. Rewriting the problem by using our notation, we get:
$$(b_2 + b_4)^2 = 100(b_1 + 2b_3 + b_5)$$
Substitute the terms:
$$(bq + bq^3)^2 = 100(b + 2bq^2 + bq^4)$$
$$b^2q^2(1 + q^2)^2 = 100b(1 + 2q^2 + q^4)$$
Divide both sides by $b$ (since $b > 0$):
$$bq^2(1 + q^2)^2 = 100(1 + 2q^2 + q^4)$$
Let $bq^2 = 100$, then $b = \frac{100}{q^2}$.

Now, since the sequence consists of natural numbers, $b$ is a natural number and $q = \frac{m}{n}$ is rational, where $(m, n) = 1$. Therefore $b m^2 = 100 n^2$. Hence we have $m \in \{2, 5, 10\}$ as $m > n$ and $m^2 \mid 100$.

Let us recall that $b_3 = 100$ for all cases.

For $q = 2$:
$$b_4 = bq^3 = 100 \cdot 2 / 4 = 200,\quad b_5 = bq^4 = 100 \cdot 4 / 4 = 400.$$
For $q = 5$:
$$b_4 = bq^3 = 100 \cdot 125 / 25 = 500,\quad b_5 = bq^4 = 100 \cdot 625 / 25 = 2500.$$
For $q = 10$:
$$b_4 = bq^3 = 100 \cdot 1000 / 100 = 1000,\quad b_5 = bq^4 = 100 \cdot 10000 / 100 = 10000.$$
For $q = \frac{5}{2}$:
$$b_4 = bq^3 = 100 \cdot \left(\frac{5}{2}\right)^3 / \left(\frac{5}{2}\right)^2 = 100 \cdot \frac{125}{8} / \frac{25}{4} = 100 \cdot 5 = 250,\quad b_5 = bq^4 = 100 \cdot \left(\frac{5}{2}\right)^4 / \left(\frac{5}{2}\right)^2 = 100 \cdot \frac{625}{16} / \frac{25}{4} = 100 \cdot 25 = 625.$$

So the desired value is $625$.