Answer: (a)(i) \(S_n=n(n+2)\sin2x\); (a)(ii) \(-220\sqrt3\); (b)(i) \(2^{n-1}\ln(2y)\); (b)(ii) \((2^n-1)\ln(2y)\); (c) \(-\dfrac38\lt w\lt \dfrac58\).

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.

The arithmetic progression has first term

\(3\sin2x\)

and common difference

\(2\sin2x.\)

Using

\(S_n=\frac n2\{2a+(n-1)d\}\),

we get

\(S_n=\frac n2\{2(3\sin2x)+(n-1)(2\sin2x)\}.\)

So

\(S_n=\frac n2\{6\sin2x+2(n-1)\sin2x\}.\)

This simplifies to

\(S_n=\frac n2(2n+4)\sin2x=n(n+2)\sin2x.\)

Therefore \(a=2\) in the required form.

When \(x=\frac{2\pi}{3}\),

\(\sin2x=\sin\frac{4\pi}{3}=-\frac{\sqrt3}{2}.\)

The sum of the first \(20\) terms is

\(S_{20}=20(20+2)\left(-\frac{\sqrt3}{2}\right).\)

Hence

\(S_{20}=-220\sqrt3.\)

For the geometric progression, the first term is \(\ln(2y)\).

The second term is

\(\ln(4y^2)=\ln((2y)^2)=2\ln(2y).\)

The third term is

\(\ln(16y^4)=\ln((2y)^4)=4\ln(2y).\)

So the common ratio is \(2\).

The \(n\)th term is therefore

\(2^{n-1}\ln(2y).\)

The sum to \(n\) terms is

\(\ln(2y)(1+2+2^2+\cdots+2^{n-1}).\)

Using the geometric sum formula,

\(S_n=\ln(2y)\frac{2^n-1}{2-1}.\)

Therefore

\(S_n=(2^n-1)\ln(2y).\)

For part (c), the first three terms are

\(\left(2w-\frac14\right),\quad\left(2w-\frac14\right)^2,\quad\left(2w-\frac14\right)^3.\)

This is a geometric progression with common ratio

\(2w-\frac14.\)

A geometric progression has a sum to infinity when the common ratio \(r\) satisfies \(|r|\lt 1\).

So

\(\left|2w-\frac14\right|\lt 1.\)

This means

\(-1\lt 2w-\frac14\lt 1.\)

Add \(\frac14\) throughout:

\(-\frac34\lt 2w\lt \frac54.\)

Divide by \(2\):

\(-\frac38\lt w\lt \frac58.\)

This completes the solution and gives the required result.