\(The first arithmetic progression (AP1) has first term 15 and common difference 4 (since 19 - 15 = 4).\)

The sum of the first n terms of an arithmetic progression is given by:

\(S_n = \frac{n}{2} [2a + (n-1)d]\)

For AP1, \(a = 15\) and \(d = 4\), so:

\(S_{n1} = \frac{n}{2} [2 \times 15 + (n-1) \times 4]\)

\(S_{n1} = \frac{n}{2} [30 + 4n - 4]\)

\(S_{n1} = \frac{n}{2} [4n + 26]\)

\(The second arithmetic progression (AP2) has first term 420 and common difference -5 (since 415 - 420 = -5).\)

For AP2, \(a = 420\) and \(d = -5\), so:

\(S_{n2} = \frac{n}{2} [2 \times 420 + (n-1) \times (-5)]\)

\(S_{n2} = \frac{n}{2} [840 - 5n + 5]\)

\(S_{n2} = \frac{n}{2} [845 - 5n]\)

Since the sums are equal, \(S_{n1} = S_{n2}\):

\(\frac{n}{2} [4n + 26] = \frac{n}{2} [845 - 5n]\)

Cancel \(\frac{n}{2}\) from both sides:

\(4n + 26 = 845 - 5n\)

\(9n = 819\)

\(n = 91\)