Answer: (a)(i) \(d=-\frac2{3p}\); (a)(ii) \(k=-5\); (b) \(r=\frac34\) or \(r=\frac14\).

(a)(i) Since the first three terms form an arithmetic progression, the common difference is the same between consecutive terms:

\(\frac1q-\frac1p=-\frac1q-\frac1q.\)

So

\(\frac1q-\frac1p=-\frac2q.\)

Rearrange:

\(\frac3q=\frac1p.\)

Hence

\(q=3p.\)

The common difference is

\(d=\frac1q-\frac1p.\)

Substitute \(q=3p\):

\(d=\frac1{3p}-\frac1p=-\frac2{3p}.\)

(a)(ii) The \(10\)th term is

\(\frac1p+9d.\)

Using \(d=-\frac2{3p}\),

Therefore

\(k=-5.\)

(b) Let the first term be \(a\) and the common ratio be \(r\).

The second term is

\(ar=\frac32.\)

The sum to infinity is

\(\frac{a}{1-r}=8.\)

From \(ar=\frac32\),

\(a=\frac{3}{2r}.\)

Substitute into the sum formula:

\(\frac{\frac{3}{2r}}{1-r}=8.\)

Thus

\(3=16r(1-r).\)

So

\(16r^2-16r+3=0.\)

Solving,

\(r=\frac34 \quad\text{or}\quad r=\frac14.\)