The binomial expansion of \((a + bx)^7\) gives the general term as \(\binom{7}{r} a^{7-r} (bx)^r\).

The coefficient of \(x\) is \(\binom{7}{1} a^6 b = 7a^6b\).

The coefficient of \(x^2\) is \(\binom{7}{2} a^5 b^2 = 21a^5b^2\).

The coefficient of \(x^4\) is \(\binom{7}{4} a^3 b^4 = 35a^3b^4\).

These coefficients form a geometric progression, so:

\(\frac{21a^5b^2}{7a^6b} = \frac{35a^3b^4}{21a^5b^2}\)

Simplifying the left side:

\(\frac{21a^5b^2}{7a^6b} = \frac{3b}{a}\)

Simplifying the right side:

\(\frac{35a^3b^4}{21a^5b^2} = \frac{5b^2}{3a^2}\)

Equating the two expressions:

\(\frac{3b}{a} = \frac{5b^2}{3a^2}\)

Cross-multiplying gives:

\(9ab^2 = 5a^3\)

Dividing both sides by \(ab^2\):

\(9 = 5 \frac{a^2}{b^2}\)

Thus, \(\frac{a^2}{b^2} = \frac{9}{5}\).

Taking the square root gives:

\(\frac{a}{b} = \frac{3}{\sqrt{5}} = \frac{5}{9}\)