(a) For a geometric progression, the ratio between consecutive terms is constant. Therefore, we have:

\(\frac{5p}{2p + 6} = \frac{8p + 2}{5p}\)

Cross-multiplying gives:

\(5p(5p) = (2p + 6)(8p + 2)\)

\(25p^2 = 16p^2 + 12p + 48p + 12\)

\(25p^2 = 16p^2 + 60p + 12\)

Rearranging gives:

\(9p^2 - 60p - 12 = 0\)

Factoring gives:

\((9p + 2)(p - 6) = 0\)

Thus, \(p = \frac{-2}{9}\) or \(p = 6\).

(b) Using \(p = \frac{-2}{9}\), the first term \(a\) is:

\(a = 2\left(\frac{-2}{9}\right) + 6 = \frac{50}{9}\)

The common ratio \(r\) is:

\(r = \frac{5p}{2p + 6} = \frac{10/9}{50/9} = \frac{-1}{5}\)

The sum to infinity \(S_\infty\) is given by:

\(S_\infty = \frac{a}{1 - r} = \frac{50/9}{1 - (-1/5)} = \frac{50/9}{6/5} = \frac{125}{27}\)

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

The second term is given by \(ar = 54\).

The sum to infinity is given by \(\frac{a}{1-r} = 243\).

From \(ar = 54\), we have \(a = \frac{54}{r}\).

Substitute \(a\) in the sum to infinity equation:

\(\frac{54}{r(1-r)} = 243\)

\(54 = 243r(1-r)\)

\(243r^2 - 243r + 54 = 0\)

Divide through by 9:

\(27r^2 - 27r + 6 = 0\)

Solving this quadratic equation using the quadratic formula:

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

\(r = \frac{27 \pm \sqrt{27^2 - 4 \times 27 \times 6}}{2 \times 27}\)

\(r = \frac{27 \pm \sqrt{729 - 648}}{54}\)

\(r = \frac{27 \pm \sqrt{81}}{54}\)

\(r = \frac{27 \pm 9}{54}\)

\(r = \frac{36}{54} = \frac{2}{3}\) or \(r = \frac{18}{54} = \frac{1}{3}\)

Since \(r > \frac{1}{2}\), we choose \(r = \frac{2}{3}\).

Substitute \(r = \frac{2}{3}\) back to find \(a\):

\(ar = 54\)

\(a \times \frac{2}{3} = 54\)

\(a = 54 \times \frac{3}{2} = 81\)

The tenth term is given by \(ar^9\):

\(ar^9 = 81 \left( \frac{2}{3} \right)^9\)

\(ar^9 = 81 \times \frac{512}{19683}\)

\(ar^9 = \frac{81 \times 512}{19683}\)

\(ar^9 = \frac{512}{243}\)

Let the first term of the geometric progression be \(a\) and the common ratio be \(r\).

The second term is \(ar\).

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

According to the problem, \(ar = 0.24 \times \frac{a}{1-r}\).

Thus, we have:

\(ar = \frac{24}{100} \times \frac{a}{1-r}\)

\(ar = \frac{24a}{100(1-r)}\)

\(100ar(1-r) = 24a\)

\(100ar - 100ar^2 = 24a\)

Divide through by \(a\) (assuming \(a \neq 0\)):

\(100r - 100r^2 = 24\)

\(100r^2 - 100r + 24 = 0\)

Factor the quadratic equation:

\((20r - 8)(5r - 3) = 0\)

Setting each factor to zero gives:

\(20r - 8 = 0 \quad \Rightarrow \quad r = \frac{8}{20} = \frac{2}{5}\)

\(5r - 3 = 0 \quad \Rightarrow \quad r = \frac{3}{5}\)

Thus, the possible values of the common ratio are \(r = \frac{2}{5}\) and \(r = \frac{3}{5}\).

The binomial expansion of \((a + bx)^7\) gives the general term as \(\binom{7}{r} a^{7-r} (bx)^r\).

The coefficient of \(x\) is \(\binom{7}{1} a^6 b = 7a^6b\).

The coefficient of \(x^2\) is \(\binom{7}{2} a^5 b^2 = 21a^5b^2\).

The coefficient of \(x^4\) is \(\binom{7}{4} a^3 b^4 = 35a^3b^4\).

These coefficients form a geometric progression, so:

\(\frac{21a^5b^2}{7a^6b} = \frac{35a^3b^4}{21a^5b^2}\)

Simplifying the left side:

\(\frac{21a^5b^2}{7a^6b} = \frac{3b}{a}\)

Simplifying the right side:

\(\frac{35a^3b^4}{21a^5b^2} = \frac{5b^2}{3a^2}\)

Equating the two expressions:

\(\frac{3b}{a} = \frac{5b^2}{3a^2}\)

Cross-multiplying gives:

\(9ab^2 = 5a^3\)

Dividing both sides by \(ab^2\):

\(9 = 5 \frac{a^2}{b^2}\)

Thus, \(\frac{a^2}{b^2} = \frac{9}{5}\).

Taking the square root gives:

\(\frac{a}{b} = \frac{3}{\sqrt{5}} = \frac{5}{9}\)