Answer: (a) \(20\) terms; (b)(i) first term \(3^{-3}\), common ratio \(3\); (b)(ii) \(3^{26}\); (c) \(|\sin\theta|\lt 1\).

(a) The arithmetic progression has first term

\(a=-4\)

and common difference

\(d=8-(-4)=12.\)

The sum of the first \(n\) terms is

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

So

We need

\(6n^2-10n\gt 2000.\)

Testing the boundary values,

\(S_{19}=6(19)^2-10(19)=1976\)

and

\(S_{20}=6(20)^2-10(20)=2200.\)

Therefore the smallest number of terms is

\(20.\)

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

The \(7\)th and \(9\)th terms give

\(ar^6=27,\qquad ar^8=243.\)

Dividing the second equation by the first,

\(r^2=9.\)

The common ratio is positive, so

\(r=3.\)

Now

\(a3^6=27=3^3,\)

so

\(a=3^{-3}.\)

(b)(ii) The \(30\)th term is

\(ar^{29}=3^{-3}\cdot3^{29}=3^{26}.\)

(c) For

\(-\frac{\pi}{2}\lt \theta\lt \frac{\pi}{2},\)

we have

\(-1\lt \sin\theta\lt 1.\)

So the common ratio of the geometric progression has magnitude less than \(1\):

\(|\sin\theta|\lt 1.\)

Therefore the progression has a sum to infinity.