Answer: \(S_n=\frac{n}{2}(3n-1)\log_x3\); \(0\lt\theta\lt\frac{\pi}{6}\).

Apply the laws of logarithms carefully, paying attention to the base of each logarithm and any restrictions on the argument.

The first three terms are

\(\log_x3,\quad \log_x81,\quad \log_x2187.\)

Since \(81=3^4\) and \(2187=3^7\), these are

\(\log_x3,\quad 4\log_x3,\quad 7\log_x3.\)

So the common difference is \(3\log_x3\).

The sum to \(n\) terms is

\(S_n=\frac{n}{2}\left(2\log_x3+(n-1)3\log_x3\right).\)

Hence

\(S_n=\frac{n}{2}(3n-1)\log_x3.\)

For the geometric progression, the common ratio is

\(r=\frac{3\tan^2\theta}{1}=3\tan^2\theta.\)

A geometric progression has a sum to infinity when \(|r|\lt1\). Since \(0\lt\theta\lt\frac{\pi}{2}\), \(r\gt0\), so

\(3\tan^2\theta\lt1.\)

Thus \(\tan\theta\lt\frac{1}{\sqrt3}\).

In the given interval, this gives \(0\lt\theta\lt\frac{\pi}{6}\).

This gives the required answer: \(S_n=\frac{n}{2}(3n-1)\log_x3\); \(0\lt\theta\lt\frac{\pi}{6}\).