Answer: (a)(i) week \(29\); (a)(ii) \(150\text{ km}\); (b)(i) \(1.25\); (b)(ii) week \(11\); (b)(iii) \(198\text{ km}\).

(a)(i) Jess's run distances form an arithmetic progression with first term \(2\) and common difference \(0.5\). The \(n\)th term is

\(2+(n-1)(0.5).\)

Set this equal to \(16\):

\(2+(n-1)(0.5)=16.\)

So

\(n-1=28,\)

and \(n=29\).

(a)(ii) The total distance for one run on each of the first 8 weeks is

\(\displaystyle \frac82(2(2)+7(0.5)).\)

Jess runs 5 times each week, so the total distance is

\(5\cdot\frac82(4+3.5)=150.\)

Thus Jess has run \(150\text{ km}\).

(b)(i) Kyle's distances form a geometric progression. Since

\(\displaystyle \frac{2.5}{2}=1.25,\)

the common ratio is \(1.25\).

(b)(ii) Kyle's distance in week \(n\) is

\(2(1.25)^{n-1}.\)

We need

\(2(1.25)^{n-1}\gt 16.\)

So

\((1.25)^{n-1}\gt 8.\)

Taking logarithms,

The first integer value is \(n=11\), so the first week is week \(11\).

(b)(iii) The sum of the first 8 weekly run distances is

\(\displaystyle \frac{2(1.25^8-1)}{1.25-1}.\)

Since Kyle runs 5 times each week, the total distance is

\(5\cdot\frac{2(1.25^8-1)}{1.25-1}=198.36\ldots.\)

So the total distance is \(198\text{ km}\) to 3 significant figures.