Answer: (a)(i) \(A=3\), \(B=-1\); (a)(ii) \(n=37\); (a)(iii) \(x=-\frac14\); (b) \(y=-\frac5{12}\) or \(y=-\frac7{12}\); (c) \(0\lt\theta\lt\frac{\pi}{4}\).

Use the appropriate formula for an arithmetic or geometric sequence, and check the common difference or common ratio before substituting.

(a)(i) The first term is

\((2x+1),\)

and the common difference is

\(4(2x+1)-(2x+1)=3(2x+1).\)

The sum to \(n\) terms is

\(S_n=\frac n2\left(2(2x+1)+(n-1)3(2x+1)\right).\)

Factorising \((2x+1)\),

\(S_n=\frac n2(2x+1)(2+3n-3).\)

So

\(S_n=\frac n2(2x+1)(3n-1).\)

Hence

(a)(ii) Given

\(\frac n2(2x+1)(3n-1)=(54n+37)(2x+1),\)

and \(x\ne-\frac12\), cancel \(2x+1\):

\(\frac n2(3n-1)=54n+37.\)

So

\(3n^2-n=108n+74.\)

Hence

\(3n^2-109n-74=0.\)

Solving gives

\(n=37\)

as the positive integer solution.

(a)(iii) With \(n=37\),

\(54n+37=54(37)+37=2035.\)

The sum is \(1017.5=\frac{2035}{2}\), so

\(2035(2x+1)=\frac{2035}{2}.\)

Therefore

\(2x+1=\frac12.\)

So

\(x=-\frac14.\)

(b) The common ratio of the geometric progression is

\(r=\frac{3(2y+1)^2}{2y+1}=3(2y+1).\)

Since the \(n\)th term is equal to 4 times the \((n+2)\)th term,

\(u_n=4u_{n+2}.\)

But \(u_{n+2}=u_nr^2\), so

\(u_n=4u_nr^2.\)

Since \(y\ne-\frac12\), \(u_n\ne0\), and therefore

\(1=4r^2.\)

Thus

\(r=\pm\frac12.\)

So

\(3(2y+1)=\pm\frac12.\)

If \(3(2y+1)=\frac12\), then \(2y+1=\frac16\), giving

\(y=-\frac5{12}.\)

If \(3(2y+1)=-\frac12\), then \(2y+1=-\frac16\), giving

\(y=-\frac7{12}.\)

(c) The common ratio is

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

For a sum to infinity,

\(\lvert r\rvert\lt1.\)

Since \(0\lt\theta\lt\frac{\pi}{2}\), \(\sin\theta\gt0\), so

\(2\sin^2\theta\lt1.\)

Hence

\(\sin\theta\lt\frac1{\sqrt2}.\)

In the interval \(0\lt\theta\lt\frac{\pi}{2}\), this gives

\(0\lt\theta\lt\frac{\pi}{4}.\)