Let the first term be \(a\) and the common ratio be \(r\).

The second term is \(ar\) and the third term is \(ar^2\).

According to the problem, \(ar = a - 9\), so:

\(a - ar = 9\)

Also, \(ar + ar^2 = 30\), so:

\(ar + ar^2 = 30\)

From \(a - ar = 9\), we have:

\(a(1 - r) = 9\)

From \(ar + ar^2 = 30\), we have:

\(ar(1 + r) = 30\)

Eliminating \(a\) from these equations, we substitute \(a = \frac{9}{1-r}\) into the second equation:

\(\frac{9r(1 + r)}{1 - r} = 30\)

Solving for \(r\), we get:

\(9r + 9r^2 = 30 - 30r\)

\(9r^2 + 39r - 30 = 0\)

Dividing by 3:

\(3r^2 + 13r - 10 = 0\)

Solving this quadratic equation, we find:

\(r = \frac{2}{3}\)

Substituting \(r = \frac{2}{3}\) back into \(a(1 - r) = 9\):

\(a \left(1 - \frac{2}{3}\right) = 9\)

\(a \cdot \frac{1}{3} = 9\)

\(a = 27\)

Thus, the first term is 27.