The sum of the first \(n\) terms of a geometric progression is given by:

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

The sum to infinity is:

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

According to the problem, \(S_n < 0.9 S_\infty\), so:

\(\frac{a(1-r^n)}{1-r} < 0.9 \frac{a}{1-r}\)

Cancel \(\frac{a}{1-r}\) from both sides:

\(1-r^n < 0.9\)

Rearrange to find:

\(r^n > 0.1\)