(i) Show \( p = \pm \frac{1}{\sqrt{3}} \):

The series is \( \ln x, p \ln x, \frac{1}{3} \ln x, \dots \). For a geometric series, the ratio between consecutive terms is constant. First term: \( a = \ln x \), second term: \( p \ln x \), third term: \( \frac{1}{3} \ln x \).

Common ratio \( r \): \( \frac{p \ln x}{\ln x} = p \), and \( \frac{\frac{1}{3} \ln x}{p \ln x} = \frac{1}{3p} \).

For geometric series: \( p = \frac{1}{3p} \).

\[ p^2 = \frac{1}{3} \]

\[ p = \pm \sqrt{\frac{1}{3}} = \pm \frac{1}{\sqrt{3}} \]

(ii) Find \( x \) given \( p > 0 \) and \( s_\infty = 3 + \sqrt{3} \):

Since \( p > 0 \), use \( p = \frac{1}{\sqrt{3}} \). The series is geometric with first term \( \ln x \) and common ratio \( p = \frac{1}{\sqrt{3}} \).

Sum to infinity: \( s_\infty = \frac{a}{1 – r} = \frac{\ln x}{1 – \frac{1}{\sqrt{3}}} \).

\[ 1 – \frac{1}{\sqrt{3}} = \frac{\sqrt{3} – 1}{\sqrt{3}} \]

\[ s_\infty = \frac{\ln x}{\frac{\sqrt{3} – 1}{\sqrt{3}}} = \frac{\sqrt{3} \ln x}{\sqrt{3} – 1} \]

Given \( s_\infty = 3 + \sqrt{3} \):

\[ \frac{\sqrt{3} \ln x}{\sqrt{3} – 1} = 3 + \sqrt{3} \]

\[ \ln x = \frac{(3 + \sqrt{3})(\sqrt{3} – 1)}{\sqrt{3}} = \frac{3\sqrt{3} – 3 + 3 – \sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{\sqrt{3}} = 2 \]

\[ x = e^2 \]

Answer: \( x = e^2 \)

(i) Show \( p = \frac{2}{3} \):

For an arithmetic series, the difference between consecutive terms is constant. Terms: \( \ln x, p \ln x, \frac{1}{3} \ln x \).

Common difference \( d \): \( p \ln x – \ln x = (p – 1) \ln x \), and \( \frac{1}{3} \ln x – p \ln x = \left( \frac{1}{3} – p \right) \ln x \).

\[ (p – 1) \ln x = \left( \frac{1}{3} – p \right) \ln x \]

Since \( \ln x \neq 0 \), divide by \( \ln x \):

\[ p – 1 = \frac{1}{3} – p \]

\[ 2p = \frac{4}{3} \]

\[ p = \frac{2}{3} \]

(ii) Write \( d \) in the form \( k \ln x \):

Using \( p = \frac{2}{3} \), common difference: \( d = (p – 1) \ln x = \left( \frac{2}{3} – 1 \right) \ln x = -\frac{1}{3} \ln x \).

\[ d = k \ln x, \quad k = -\frac{1}{3} \]

Answer: \( d = -\frac{1}{3} \ln x \)

(iii) Find \( n \) given sum is \( -3 \ln x \):

Arithmetic series with first term \( a = \ln x \), common difference \( d = -\frac{1}{3} \ln x \). Sum of first \( n \) terms:

\[ s_n = \frac{n}{2} [2a + (n – 1)d] \]

\[ = \frac{n}{2} \left[ 2 \ln x + (n – 1) \left( -\frac{1}{3} \ln x \right) \right] \]

\[ = \frac{n}{2} \ln x \left[ 2 – \frac{n – 1}{3} \right] = \ln x \cdot \frac{n}{2} \cdot \frac{6 – (n – 1)}{3} = \ln x \cdot \frac{n (7 – n)}{6} \]

Given \( s_n = -3 \ln x \):

\[ \frac{n (7 – n)}{6} \ln x = -3 \ln x \]

Divide by \( \ln x \):

\[ \frac{n (7 – n)}{6} = -3 \]

\[ n (7 – n) = -18 \]

\[ n^2 – 7n – 18 = 0 \]

Solve: Discriminant \( \Delta = (-7)^2 – 4 \cdot 1 \cdot (-18) = 49 + 72 = 121 \).

\[ n = \frac{7 \pm \sqrt{121}}{2} = \frac{7 \pm 11}{2} \]

\[ n = 9 \text{ or } n = -2 \]

Since \( n \) is a positive integer (number of terms), \( n = 9 \).

Answer: \( n = 9 \)