Answer: (a) \(a=\frac{11}{2}d\). (b) \(\frac{S_B}{S_A}=\frac{r}{1+r}\).

(a) For an arithmetic progression,

\(S_n=\dfrac n2\left(2a+(n-1)d\right).\)

So

\(S_{20}=10(2a+19d),\)

and

\(S_{10}=5(2a+9d).\)

It is given that \(S_{20}=3S_{10}\), so

\(10(2a+19d)=3\cdot5(2a+9d).\)

Expanding,

\(20a+190d=30a+135d.\)

Therefore

\(55d=10a,\)

so

\(a=\dfrac{11}{2}d.\)

(b) Progression A has first term \(a_1\) and common ratio \(r\). Since \(|r|\lt1\), its sum to infinity is

\(S_A=\dfrac{a_1}{1-r}.\)

The terms of progression B are

\(a_2,\ a_4,\ a_6,\ldots\)

Since \(a_2=a_1r\), the first term of B is \(a_1r\). Moving from \(a_2\) to \(a_4\) multiplies by \(r^2\), so the common ratio of B is \(r^2\).

Therefore

\(S_B=\dfrac{a_1r}{1-r^2}.\)

Now

\(\dfrac{S_B}{S_A}=\dfrac{\frac{a_1r}{1-r^2}}{\frac{a_1}{1-r}}.\)

So

\(\dfrac{S_B}{S_A}=\dfrac{a_1r}{1-r^2}\cdot\dfrac{1-r}{a_1}.\)

Since \(1-r^2=(1-r)(1+r)\), this simplifies to

\(\dfrac{S_B}{S_A}=\dfrac{r}{1+r}.\)