The formula for the sum of the first n terms of a geometric progression is given by:

\(S_n = \frac{a(1-r^n)}{1-r}\)

where \(a\) is the first term and \(r\) is the common ratio.

The formula for the sum to infinity of a geometric progression is:

\(S_\infty = \frac{a}{1-r}\)

Given \(r = 0.99\), we need to find \(\frac{S_{100}}{S_\infty} \times 100\).

First, calculate \(1 - r^n\) for \(n = 100\):

\(1 - 0.99^{100} \approx 0.634\)

Then, the ratio \(\frac{S_{100}}{S_\infty}\) is:

\(\frac{a(1-0.99^{100})}{a} = 1 - 0.99^{100} \approx 0.634\)

Expressing this as a percentage:

\(0.634 \times 100 \approx 63.4\%\)

Rounding to 2 significant figures gives:

63%