Answer: \(t=-1.5\), and the common ratio is \(3\).

For an arithmetic progression with first term \(t\) and common difference \(1.5\), the \(n\)th term is

\(t+(n-1)(1.5)\).

Therefore

\(4\text{th term}=t+3(1.5)=t+4.5\),

\(8\text{th term}=t+7(1.5)=t+10.5\),

and

\(20\text{th term}=t+19(1.5)=t+28.5\).

These three terms are consecutive terms of a geometric progression. For three consecutive terms of a geometric progression, the square of the middle term is equal to the product of the other two terms. Hence

\((t+10.5)^2=(t+4.5)(t+28.5)\).

Expanding both sides,

\(t^2+21t+110.25=t^2+33t+128.25\).

Subtracting \(t^2+21t+110.25\) from both sides gives

\(0=12t+18\).

Thus

\(t=-\frac{18}{12}=-1.5\).

The three terms are then

\(t+4.5=3\), \(t+10.5=9\), \(t+28.5=27\).

So the common ratio of the geometric progression is

\(\dfrac{9}{3}=3\).

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

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

Answer: (a) \(d=6\). (b) \(r=3\).

(a) For an arithmetic progression with first term \(3\) and common difference \(d\),

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

Thus

\(S_{10}=\dfrac{10}{2}(6+9d)=5(6+9d),\)

and

\(S_5=\dfrac52(6+4d).\)

It is given that \(S_{10}=4S_5\), so

\(5(6+9d)=4\cdot\dfrac52(6+4d).\)

Expanding,

\(30+45d=60+40d.\)

Therefore

\(5d=30,\)

so

\(d=6.\)

(b) Let the first term of the arithmetic progression be \(a\), and let its common difference be \(d\). Then its first, second and fifth terms are

\(a,\quad a+d,\quad a+4d.\)

These are the first three terms of a geometric progression. If the common ratio is \(r\), then

\(a+d=ar,\qquad a+4d=ar^2.\)

From the first equation,

\(d=a(r-1).\)

Substitute this into the second equation:

\(a+4a(r-1)=ar^2.\)

Since the first term is non-zero, \(a\ne0\), so divide by \(a\):

\(1+4(r-1)=r^2.\)

Hence

\(r^2-4r+3=0.\)

Factorising,

\((r-1)(r-3)=0.\)

The common ratio is not \(1\), so

\(r=3.\)