(a) Let the first term be \(a\) and the common ratio be \(r\). The second term is given by \(ar = 16\). The sum to infinity is given by \(\frac{a}{1-r} = 100\).

From \(ar = 16\), we have \(a = \frac{16}{r}\).

From \(\frac{a}{1-r} = 100\), we have \(a = 100(1-r)\).

Equating the two expressions for \(a\):

\(\frac{16}{r} = 100(1-r)\)

\(16 = 100r - 100r^2\)

\(100r^2 - 100r + 16 = 0\)

Solving this quadratic equation using the quadratic formula:

\(r = \frac{5 \pm \sqrt{25 - 4 \times 1 \times 4}}{2 \times 5}\)

\(r = \frac{5 \pm 3}{10}\)

\(r = \frac{4}{5}\) or \(r = \frac{1}{5}\)

For \(r = \frac{4}{5}\), \(a = \frac{16}{\frac{4}{5}} = 20\).

For \(r = \frac{1}{5}\), \(a = \frac{16}{\frac{1}{5}} = 80\).

Thus, the two possible values of the first term are 20 and 80.

(b) For \(r = \frac{4}{5}\), the nth term is \(u_n = 20 \left(\frac{4}{5}\right)^{n-1}\).

For \(r = \frac{1}{5}\), the nth term is \(v_n = 80 \left(\frac{1}{5}\right)^{n-1}\).

We need to show \(u_n = 4^{n-2} \times v_n\).

\(u_n = 20 \left(\frac{4}{5}\right)^{n-1} = 20 \times \frac{4^{n-1}}{5^{n-1}}\)

\(v_n = 80 \left(\frac{1}{5}\right)^{n-1} = 80 \times \frac{1}{5^{n-1}}\)

\(u_n = 4^{n-2} \times v_n\) simplifies to:

\(20 \times \frac{4^{n-1}}{5^{n-1}} = 4^{n-2} \times 80 \times \frac{1}{5^{n-1}}\)

\(20 \times 4^{n-1} = 4^{n-2} \times 80\)

\(4^{n-1} = 4^{n-2} \times 4\)

Thus, the nth term of one progression is equal to \(4^{n-2}\) times the nth term of the other progression.