Answer: (a)(ii) \(a=2\). (b) \(p=-8\), \(q=12\).

Answer: (a)(ii) \(a=2\). (b) \(p=-8\), \(q=12\).

(a)(i) In the geometric progression, the second term is \(ar\), so

\(ar=8\).

The third and fourth terms are \(ar^2\) and \(ar^3\), and their sum is 160:

\(ar^2+ar^3=160\).

Since \(ar=8\), write \(a=\frac8r\). Then

\(\frac8r r^2+\frac8r r^3=160\).

So

\(8r+8r^2=160\).

Dividing by 8 gives

\(r+r^2=20\),

or

\(r^2+r-20=0\), as required.

(a)(ii) Factorise:

\((r+5)(r-4)=0\).

Since \(r\gt 0\),

\(r=4\).

Using \(ar=8\),

\(4a=8\),

so

\(a=2\).

(b) For the first arithmetic progression,

\(q\text{th term}=p+2(q-1)\).

It is given that this is 14, so

\(p+2(q-1)=14\). \(\quad (1)\)

For the second arithmetic progression, the sum of the first \(q\) terms is

\(\displaystyle \frac{q}{2}\{2p+(q-1)4\}\).

This is 168, so

\(\displaystyle \frac{q}{2}\{2p+4(q-1)\}=168\). \(\quad (2)\)

From (1),

\(p=14-2(q-1)=16-2q\).

Substitute this into (2):

\(\displaystyle \frac{q}{2}\{2(16-2q)+4(q-1)\}=168\).

Simplifying inside the brackets,

\(2(16-2q)+4(q-1)=32-4q+4q-4=28\).

Thus

\(\displaystyle \frac{q}{2}\cdot28=168\),

so

\(14q=168\).

Hence

\(q=12\).

Then

\(p=16-2(12)=-8\).