Answer: (a) first term \(-16\), \(21\)st term \(24\); (b)(i) \(r=\frac{p}{3}\); (b)(ii) \(S_\infty=\frac{243p}{3-p}\); (b)(iii) \(p=\frac34\).

(a) Let the first term be \(a\) and the common difference be \(d\).

The second term is \(-14\), so

\(a+d=-14.\)

The sum to \(21\) terms is

\(S_{21}=\frac{21}{2}(2a+20d)=84.\)

Divide by \(21\):

\(\frac12(2a+20d)=4.\)

So

\(a+10d=4.\)

Subtract \(a+d=-14\) from \(a+10d=4\):

\(9d=18.\)

Hence \(d=2\). Then

\(a+2=-14,\)

so

\(a=-16.\)

The \(21\)st term is

\(a+20d=-16+40=24.\)

(b)(i) Let the first term of the geometric progression be \(A\).

The second term is

\(Ar=27p^2,\)

and the fifth term is

\(Ar^4=p^5.\)

Divide the fifth-term equation by the second-term equation:

\(\frac{Ar^4}{Ar}=\frac{p^5}{27p^2}.\)

So

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

Since \(0\lt r\lt1\),

\(r=\frac{p}{3}.\)

(b)(ii) From \(Ar=27p^2\),

\(A=\frac{27p^2}{r}.\)

Using \(r=\frac{p}{3}\),

\(A=\frac{27p^2}{p/3}=81p.\)

The sum to infinity is

\(S_\infty=\frac{A}{1-r}.\)

Therefore

\(S_\infty=\frac{81p}{1-\frac{p}{3}} =\frac{243p}{3-p}.\)

(b)(iii) Given that \(S_\infty=81\),

\(81=\frac{243p}{3-p}.\)

So

\(81(3-p)=243p.\)

Divide by \(81\):

\(3-p=3p.\)

Hence

\(4p=3,\)

and therefore

\(p=\frac34.\)