How do we get the formula of the sequence? A woman is walking home with distance $d$ and speed $v$.
The dog is happy and runs at speed $\frac{3v }{ 2}$ always between woman and House back and forth.
(i) At what distances $(d_n)_{n\geq 1}$ do women and dogs meet?
(ii) Determine the total path length of the dog with the help of $(d_n)_{n\geq 1}$.
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EDIT:
Let $x$ be the distance between the woman and the house and let $s$ be the distance between the dog and the house.
At the beginning we have $x_0=d$ and $s_0=0$.
It holds that $x=d-vt$ and $s=\frac 32v t$.
The woman and the dog meet at $x=s=d_1$ for $t=t_1$.
From $x=s$ we get $d-vt=\frac{3}{2}vt \Rightarrow \frac{5}{2}vt=d \Rightarrow t=\frac{2d}{5v}$, so $t_1=\frac{2d}{5v}$.
So we have that $d_1=x(t_1)=d-v\cdot \frac{2d}{5v}=d- \frac{2d}{5}=\frac{3d}{5}$.
After the meeting of the woman and the dog, the dog goes back to the house and then again to the woman.
At time $t_1$ the dog is at $s_1=\frac{3d}{5}$.
The dog goes in direction of the house, so $s=\frac{3d}{5}-\frac{3v}{2}t$.
When $s=0$ the dog arrived at the house, so $t=\frac{2d}{5v}$.
Then the dog goes back in direction of the woman, so $s= \frac{3}{2}vt$ with $t&gt;\frac{2d}{5v}$.
At time $t_1$ the woman is also at $x_1=\frac{3d}{5}$.
The woman continues in diection to the house, so $x=\frac{3d}{5}-vt$.
For the second meeting, it holds $x=s=d_2$.
We get \begin{equation*}\frac{3d}{5}-vt=\frac{3}{2}vt \Rightarrow \frac{3d}{5}=\frac{5}{2}vt \Rightarrow t=\frac{6d}{25v}\end{equation*} So  they meet at \begin{equation*}d_2=x(t_2)=\frac{3d}{5}-v\cdot \frac{6d}{25v}=\frac{9d}{25}\end{equation*}.
So the $n$-th meeting is at $d_n =\left (\frac{3}{5}\right )^nd$.
Is that correct so far?
For the second question do we take the sum of all the terms $d_n$ ? Or should we take twice each term $d_n$ since the dog goes back and forth?
The distance of the woman from home is given by $d-vt$, and that of the dog by $d-\frac32vt$, until it reaches it, at time $t_1=\frac23\frac dv$. It comes back to its mistress with equation
$$\frac32v(t-t_1)=\frac32vt-d.$$
The moment they meet is obtained by solving
$$\frac32vt_2-d=d-vt_2$$ or $$t_2=\frac25\frac dv,$$ at a distance $\frac35d$ of home.

Now this repeats proportionally, substituting $\frac35d$ for $d$, and we have a geometric progression of common ratio $\frac35$.
The distances of the meetings are given by the partial sums of the sequence
$$\frac25d,\frac25\frac35d,\frac25\frac9{25}d,\cdots\frac25\frac{3^n}{5^n}d,\cdots$$
which indeed converges to $d$.
The dog travels $\frac32$ times more than the woman, hence $\frac32d$.
If the woman and dog start at distance $d$ from the house and re-meet when the woman is at distance $\delta$ from the house, the woman will have walked a distance $d-\delta$ and the dog will have run a distance $d+\delta$. Now since the dog runs fifty percent faster than the woman walks, it goes fifty percent further in any given amount of time, meaning
$$d+\delta={3\over2}(d-\delta)$$
and thus $\delta={1\over5}d$. So the sequence of distances (from the house) that the woman and the dog meet up is $d_n=\left(1\over5\right)^nd$. The total distance the dog runs is
$$\begin{align}
(d+d_1)+(d_1+d_2)+(d_2+d_3)+\cdots
&amp;=d+2d\sum_{n=1}^\infty\left(1\over5\right)^n\\
&amp;=d+2d{1/5\over1-1/5}\\
&amp;=d+2d{1\over5-1}\\
&amp;=d+{2d\over4}\\
&amp;={3\over2}d
\end{align}$$
which agrees with what we said earlier: the dog goes fifty percent further than the woman.