Let the common ratio of the geometric progression be \(r\) and the common difference of the arithmetic progression be \(d\).

The terms of the geometric progression are \(a, ar, ar^2, ar^3, ar^4, \ldots\).

The terms of the arithmetic progression are \(a, a+d, a+2d, a+3d, a+4d, \ldots\).

From the problem, we have:

1. \(ar^2 = a + d\)

2. \(ar^4 = a + 5d\)

From equation 1: \(ar^2 = a + d\)

From equation 2: \(ar^4 = a + 5d\)

We can express \(ar^4\) in terms of \(ar^2\):

\(ar^4 = (ar^2)^2 = (a + d)^2\)

Equating the two expressions for \(ar^4\):

\(a(a + 5d) = (a + d)^2\)

Expanding both sides:

\(a^2 + 5ad = a^2 + 2ad + d^2\)

Simplifying gives:

\(3ad = d^2\)

\(d(3a - d) = 0\)

Since \(d \neq 0\), we have \(d = 3a\).

Substitute \(d = 3a\) into the sum formula for the arithmetic progression:

The sum of the first 20 terms is:

\(S_{20} = \frac{20}{2} [2a + 19 \times 3a]\)

\(S_{20} = 10 [2a + 57a]\)

\(S_{20} = 10 \times 59a\)

\(S_{20} = 590a\)