The sum to infinity of a geometric progression is given by \(S = \frac{a}{1-r}\), where \(a\) is the first term and \(r\) is the common ratio.

For the first progression, \(a = 1\) and \(r = r\), so the sum is \(S = \frac{1}{1-r}\).

For the second progression, \(a = 4\) and \(r = \frac{1}{4}r\), so the sum is \(S = \frac{4}{1-\frac{1}{4}r}\).

Since the sums are equal, we have:

\(\frac{1}{1-r} = \frac{4}{1-\frac{1}{4}r}\)

Cross-multiplying gives:

\(1 - \frac{1}{4}r = 4(1-r)\)

\(1 - \frac{1}{4}r = 4 - 4r\)

Rearranging gives:

\(4r - \frac{1}{4}r = 4 - 1\)

\(\frac{15}{4}r = 3\)

\(r = \frac{4}{5}\)

Substituting \(r = \frac{4}{5}\) back into the sum formula for the first progression:

\(S = \frac{1}{1-\frac{4}{5}} = \frac{1}{\frac{1}{5}} = 5\)

Thus, \(r = \frac{4}{5}\) and \(S = 5\).