Answer: \(n=12\); \(r=2\).

Identify whether the sequence is arithmetic or geometric, then use the correct formula for the required term or sum.

The weekly distances form a geometric progression with first term \(10\) and common ratio \(1.1\).

The total after \(n\) weeks is

\(S_n=\frac{10(1.1^n-1)}{1.1-1}=100(1.1^n-1).\)

We need \(100(1.1^n-1)\gt200\), so \(1.1^n\gt3\).

Taking logarithms,

\(n\gt\frac{\log3}{\log1.1}=11.526\ldots\).

Therefore the smallest possible integer value is \(n=12\).

For part (b), let the arithmetic progression have first term \(A\) and common difference \(d\).

The 1st, 3rd and 7th terms are \(A\), \(A+2d\), and \(A+6d\).

These are \(a\), \(ar\), and \(ar^2\), so

\(A=a,\quad A+2d=ar,\quad A+6d=ar^2.\)

Thus \(a(r-1)=2d\) and \(a(r^2-1)=6d\).

Dividing the second equation by the first gives

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

Since \(r\ne1\), this becomes \(r+1=3\), so \(r=2\).

This gives the required answer: \(n=12\); \(r=2\).