The sum to infinity of a geometric progression is given by the formula:

\(S = \frac{a}{1-r}\)

where \(a\) is the first term and \(r\) is the common ratio.

Given \(a = 100\) and \(S = 2000\), we have:

\(\frac{100}{1-r} = 2000\)

Solving for \(r\):

\(100 = 2000(1-r)\)

\(100 = 2000 - 2000r\)

\(2000r = 2000 - 100\)

\(2000r = 1900\)

\(r = \frac{1900}{2000} = \frac{19}{20}\)

The second term of the progression is \(ar\):

\(ar = 100 \times \frac{19}{20} = 95\)

The sum of the first \(n\) terms of a geometric progression is given by:

\(S_n = \frac{a(1-r^n)}{1-r}\)

The sum to infinity is:

\(S_\infty = \frac{a}{1-r}\)

According to the problem, \(S_n < 0.9 S_\infty\), so:

\(\frac{a(1-r^n)}{1-r} < 0.9 \frac{a}{1-r}\)

Cancel \(\frac{a}{1-r}\) from both sides:

\(1-r^n < 0.9\)

Rearrange to find:

\(r^n > 0.1\)

Let the first term be \(a = 16\) and the common ratio be \(r\).

The fourth term is given by \(ar^3 = \frac{27}{4}\).

Substitute \(a = 16\) into the equation:

\(16r^3 = \frac{27}{4}\)

Divide both sides by 16:

\(r^3 = \frac{27}{64}\)

Take the cube root of both sides:

\(r = \frac{3}{4}\)

The sum to infinity of a geometric progression is given by:

\(S = \frac{a}{1 - r}\)

Substitute \(a = 16\) and \(r = \frac{3}{4}\):

\(S = \frac{16}{1 - \frac{3}{4}}\)

\(S = \frac{16}{\frac{1}{4}}\)

\(S = 16 \times 4 = 64\)

Given the first term \(a = 12\) and the second term \(ar = -6\), we find the common ratio \(r\) by solving:

\(ar = -6 \Rightarrow 12r = -6 \Rightarrow r = -\frac{1}{2}\).

(i) The formula for the \(n\)-th term of a geometric progression is \(a_n = ar^{n-1}\). For the tenth term:

\(a_{10} = 12 \left(-\frac{1}{2}\right)^{9} = 12 \times -\frac{1}{512} = -\frac{3}{128}\).

(ii) The sum to infinity of a geometric progression is given by \(S_\infty = \frac{a}{1-r}\), provided \(|r| < 1\):

\(S_\infty = \frac{12}{1 - (-\frac{1}{2})} = \frac{12}{1 + \frac{1}{2}} = \frac{12}{\frac{3}{2}} = 8\).

(i) The sum of the first 3 terms of a geometric progression is given by:

\(a + ar + ar^2 = 35\)

Substitute \(r = -\frac{2}{3}\):

\(a + a(-\frac{2}{3}) + a(-\frac{2}{3})^2 = 35\)

\(a - \frac{2}{3}a + \frac{4}{9}a = 35\)

Combine terms:

\(a(1 - \frac{2}{3} + \frac{4}{9}) = 35\)

\(a(\frac{9}{9} - \frac{6}{9} + \frac{4}{9}) = 35\)

\(a(\frac{7}{9}) = 35\)

\(a = 35 \times \frac{9}{7}\)

\(a = 45\)

(ii) The sum to infinity of a geometric progression is given by:

\(S_\infty = \frac{a}{1-r}\)

Substitute \(a = 45\) and \(r = -\frac{2}{3}\):

\(S_\infty = \frac{45}{1 - (-\frac{2}{3})}\)

\(S_\infty = \frac{45}{1 + \frac{2}{3}}\)

\(S_\infty = \frac{45}{\frac{5}{3}}\)

\(S_\infty = 45 \times \frac{3}{5}\)

\(S_\infty = 27\)