Answer: (a)(i) \(n=56\); (a)(ii) \(\theta=0.001\); (b)(i) yes; (b)(ii) \(3^{2-n}\lg\phi\); (b)(iii) \(\phi=10\).

We start with the main method. Identify whether the sequence is arithmetic or geometric, then apply the appropriate formula for the required sum or term.

(a)(i) The first three terms are

\(\lg\theta^2,\quad \lg\theta^5,\quad \lg\theta^8.\)

Using \(\lg\theta^k=k\lg\theta\), these are

\(2\lg\theta,\quad5\lg\theta,\quad8\lg\theta.\)

So the common difference is

\(d=3\lg\theta.\)

The sum to \(n\) terms is

\(S_n=\frac n2(2a+(n-1)d).\)

Thus

\(\frac n2\left(2(2\lg\theta)+(n-1)3\lg\theta\right)=4732\lg\theta.\)

Dividing by \(\lg\theta\),

\(\frac n2(4+3n-3)=4732.\)

So

\(\frac n2(3n+1)=4732.\)

Hence

\(3n^2+n-9464=0.\)

Solving gives

\(n=56\)

as the positive solution.

(a)(ii) The sum is also \(-14196\), so

\(4732\lg\theta=-14196.\)

Therefore

\(\lg\theta=-3,\)

and hence

\(\theta=10^{-3}=0.001.\)

(b)(i) The first three terms are

\(\lg\phi^3,\quad \lg\phi,\quad \lg\phi^{1/3}.\)

These are

\(3\lg\phi,\quad \lg\phi,\quad \frac13\lg\phi.\)

The common ratio is

\(r=\frac{\lg\phi}{3\lg\phi}=\frac13.\)

Since \(|r|\lt1\), the geometric progression has a sum to infinity.

(b)(ii) The \(n\)th term is

\(3\lg\phi\left(\frac13\right)^{n-1}.\)

So

\(T_n=3^{2-n}\lg\phi.\)

(b)(iii) The twentieth term is

\(T_{20}=3^{2-20}\lg\phi=3^{-18}\lg\phi.\)

It is given that \(T_{20}=3^{-18}\), so

\(\lg\phi=1.\)

Therefore

\(\phi=10.\)

This completes the solution and gives the required result.