Answer: \(25\) terms; common ratio \(\frac34\); first term \(18\).

(a) Let the first term be \(a\) and the common difference be \(d\). The second term is \(a+d=8\), and the fourth term is \(a+3d=18\).

Subtracting the equations gives

\(2d=10,\)

so \(d=5\). Then

\(a+5=8,\)

so \(a=3\).

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

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

Using \(a=3\) and \(d=5\),

\(S_n=\frac n2\left(6+5n-5\right)=\frac n2(5n+1).\)

We need

\(\frac n2(5n+1)\gt 1560.\)

So

\(5n^2+n-3120\gt 0.\)

The positive root of \(5n^2+n-3120=0\) is approximately \(24.9\), so the least integer value is

\(n=25.\)

(b)(i) Let the first term be \(a\) and the common ratio be \(r\). Since the sum to infinity is \(72\),

\(\frac{a}{1-r}=72,\)

so

\(a=72(1-r).\)

The sum of the first three terms is

\(a+ar+ar^2=a(1+r+r^2)=\frac{333}{8}.\)

Substituting \(a=72(1-r)\),

\(72(1-r)(1+r+r^2)=\frac{333}{8}.\)

Since \((1-r)(1+r+r^2)=1-r^3\),

\(72(1-r^3)=\frac{333}{8}.\)

Thus

\(1-r^3=\frac{333}{576}=\frac{37}{64},\)

so

\(r^3=\frac{27}{64}.\)

Hence

\(r=\frac34.\)

(b)(ii) Now

\(a=72\left(1-\frac34\right)=18.\)