Inspired by toilet paper

Since, the area between the circles is equal,
$\pi r_{n^{2}} - \pi r_{n-1}^{2} = \pi r_{n+1}^{2} - \pi r_{n}^{2} \rightarrow 2r_{n}^{2} = r_{n-1}^{2} + r_{n+1}^{2}$
Thus the squares of these radii form an AP.
Also ,
$r_{1}^{2} = r_{2}^{2}-r_{1}^{2}$
Thus the common difference of this AP is $r_{1}^{2}$
$r_{n}^{2} = r_{1}^{2} + (n-1)r_{1}^{2}$
$\dfrac{r_{n}^{2}}{r_{1}^{2}} = 1 + n - 1 = n$
$\dfrac{r_{n}}{r_{1}} = \sqrt{n}$
Put n = 100 , answer = 10

$\begin{aligned} A_1 &= \pi r_1^2 \\ =A_2 &= \pi r_2^2- \pi r_1^2 \\ =A_3 &= \pi r_3^2- \pi r_2^2 \ \\ =A_4 &= \pi r_4^2- \pi r_3^2 \\ &\dots \\ &\dots \\ =A_n &= \pi r_n^2- \pi r_{n-1}^2 \\ Using \ Theorem \ on \ equal \ ratios& \\ =& \dfrac{ \pi r_1^2+\pi r_2^2- \pi r_1^2+\pi r_3^2- \pi r_2^2+\pi r_4^2- \pi r_3^2 \dots \pi r_n^2- \pi r_{n-1}^2 }{n} \\ &=\dfrac{\pi r_n^2}{n} \\ \Rightarrow \pi r_1^2 &=\dfrac{\pi r_n^2}{n} \\ \Rightarrow \dfrac{r_n}{r_1}=\sqrt{n} \end{aligned}$

Putting $n=100$ we have the ratio as $\sqrt{100}=\boxed{10.00}$ .

The circular areas increase with a constant increment, and are proportional to the squares of their radii, $r_i^2$ . Moreover, we may take $r_0 = 0$ . Thus the squared radius of a circle is directly proportional to its index: $r_i^2 \propto i.$ It follows that $\frac{r_n}{r_m} = \sqrt{\frac n m}.$ In this case, that results in $\sqrt{100/1} = \boxed{10}$ .