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

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

Substituting the given values, \(a = -12\) and \(d = 6\), we have:

\(S_n = \frac{n}{2} [-24 + (n-1)6]\)

We need \(S_n > 3000\):

\(\frac{n}{2} [-24 + 6n - 6] > 3000\)

\(\frac{n}{2} [6n - 30] > 3000\)

\(3n(2n - 5) > 3000\)

\(3n^2 - 15n - 3000 > 0\)

Solving the quadratic inequality \(3n^2 - 15n - 3000 = 0\), we find the roots:

\(n = 34.2 \text{ and } n = -29.2\)

Since \(n\) must be a positive integer, the least possible value of \(n\) is \(35\).

The circumference measurements form an arithmetic progression with the first term \(a = 5.00\) m and the second term \(a + d = 5.02\) m, where \(d\) is the common difference.

Thus, \(d = 5.02 - 5.00 = 0.02\) m.

The formula for the \(n\)-th term of an arithmetic progression is \(a_n = a + (n-1) imes d\).

We need to find the circumference 20 years after the first measurement, which is the 21st term (since the first measurement is at year 0).

\(a_{21} = 5.00 + (21-1) imes 0.02\)

\(a_{21} = 5.00 + 20 imes 0.02\)

\(a_{21} = 5.00 + 0.40\)

\(a_{21} = 5.40\) m

Given the sum of the first n terms \(S_n = \frac{1}{2}n(3n + 7)\).

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

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

Equating the two expressions for \(S_n\):

\(\frac{n}{2} (2a + (n-1)d) = \frac{1}{2}n(3n + 7)\)

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

\(2a + (n-1)d = 3n + 7\)

Substitute \(n = 1\):

\(2a = 3(1) + 7\)

\(2a = 10\)

\(a = 5\)

Substitute \(n = 2\):

\(2a + d = 3(2) + 7\)

\(2(5) + d = 13\)

\(10 + d = 13\)

\(d = 3\)

Thus, the first term is 5 and the common difference is 3.

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

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:

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

\(9n = 819\)

\(n = 91\)

The first term of the arithmetic progression is 16, and the second term is 24. Therefore, the common difference is given by:

\(d = 24 - 16 = 8\)

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

\(S_n = \frac{n}{2} \left( 2a + (n-1)d \right)\)

Substituting the known values \(a = 16\) and \(d = 8\), we have:

\(S_n = \frac{n}{2} \left( 32 + (n-1) \times 8 \right)\)

We need \(S_n > 20,000\):

\(\frac{n}{2} \left( 32 + 8n - 8 \right) > 20,000\)

\(\frac{n}{2} (24 + 8n) > 20,000\)

\(n(12 + 4n) > 20,000\)

\(4n^2 + 12n - 20,000 > 0\)

Dividing the entire inequality by 4:

\(n^2 + 3n - 5,000 > 0\)

Solving the quadratic equation \(n^2 + 3n - 5,000 = 0\) using the quadratic formula:

\(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 1\), \(b = 3\), \(c = -5,000\):

\(n = \frac{-3 \pm \sqrt{3^2 - 4 \times 1 \times (-5,000)}}{2 \times 1}\)

\(n = \frac{-3 \pm \sqrt{9 + 20,000}}{2}\)

\(n = \frac{-3 \pm \sqrt{20,009}}{2}\)

Approximating the square root:

\(n \approx \frac{-3 + 141.42}{2}\)

\(n \approx 69.21\)

Since \(n\) must be an integer, we round up to the nearest whole number:

\(n = 70\)

Therefore, 70 terms are needed for the sum to exceed 20,000.