**Problem:** Let \({C_n}\) be an infinite sequence of circles lying in the positive quadrant of the \(XY\)-plane, with strictly decreasing radii and satisfying the following conditions. Each \(C_n\) touches both the \(X\)-axis and the \(Y\)-axis. Further, for all \(n\geq 1\), the circle \(C_{n+1}\) touches the circle \(C_n\) externally. If \(C_1\) has radius \(10\: cm\), then show that the sum of the areas of all these circles is \(\frac{25\pi}{3\sqrt{2}-4} \: cm^2\).

**Solution:** Consider the following diagram where the Green line segment is \(R_n\), the radius of the \(n^{th}\) circle, and the Yellow line segment is \(R_{n+1}\).

As we are told about the symmetricity of the figure in the problem we can say that:

\(\sqrt{2}R_{n+1} + R_{n+1} + R_n = \sqrt{2} R_n\)

\(=> R_{n+1}(\sqrt{2}+1)=R_n(\sqrt{2}-1)\)

\(=> R_{n+1}= (3-2\sqrt{2})R_n\)

Let’s say \(=> R_{n+1}= \alpha.R_n\).

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