Is there a formula or way to plot to find x, which is the same for two different sequences please bear with me as I am self taught and haven't yet mastered notation. 
Here is the situation and I am looking for one or both ways to do this, either a formula that will give me the correct result or a way to plot this out on a graph so I can get a good idea where the answer is headed.
Situation; I have two, lets for arguments sake say cars. They are any given number of feet away from each other to start. That is one variable. When they start to move they are going to start dropping markers on the road, in segments. Car starts with a shorter segment. Their respective segments are also variable, but here's an example. When they begin car A will first drop a marker 5 feet away from its starting position, car B will drop a marker 9 feet away from it's starting position. Now when they both drop their next markers it's will be the same distance as before with an additional number of feet - which will be the same for both. So if the number of additional feet is 2 feet, car A will now drop a marker 7 feet from where the first marker was dropped and car B will drop it 11 feet instead of 9 feet. In the next segment car A will drop it's marker 9 feet away and car B will drop it 13 feet away. Eventually, at some point both cars will drop a marker at the exact same place. This is what I am trying to figure out. Is there a formula that, once the main numbers are all plugged in, which will tell me at what segment both markers will be at the same place without having to add up each cars segments continuously until I reach that number? Or, is there a way to plot it so I can see where the lines are converging towards each other so I can get a pretty good idea where that same segment point is?
Thank you.
 A: Let's formalize the question, to see if somebody has something to add.
You can assume one of the cars starts from point $0$. If the initial step is $a$ and the increment of each step is $s$, it is easy to see that the positions after $1$, $2$, $3$,... steps are
$$
a,\ \ \ 2a+s,\ \ \ 3a+3s\,\ \ \ 4a+6s,\ \ \ 5a+10s,\dots
$$
The coefficients of $s$ are just $1$, $1+2$, $1+2+3$, and so on, i.e., the triangular numbers. So after $n$ steps the car will be distance 
$$f(n)=n\cdot a+\frac{(n-1)n}{2}s =n\left(a+\frac{s(n-1)}{2}\right).$$
We can image the the other cars starts at distance $d$ and its initial step is $b$. So its position afte $m$ steps will be
$$g(m)=d+m\cdot b+\frac{(m-1)m}{2}s =d+m\left(b+\frac{s(m-1)}{2}\right).$$
So you are asking if, given $a$, $b$, $s$ and $d$ there are two values $n$ and $m$ such that $f(n)=g(m)$.
If we require that the two cars leave the mark in the same position at the same time, then the equation becomes $f(n)=g(n)$ and it is easy to solve,
giving
$$
n = \frac{d}{a-b}
$$
So if $n=d/(a-b)$ is a positive integer, you know that at the $n$-th step the two cars leave a mark in the same point. If this value is not integer, or negative, then you know that there is not such moment.
If you are confronting only the positions of the marks and not the time the marks have been made, then you have the general problem $f(n)=g(m)$.
This is a quadratic Diophantine equation (Diophantine means we are only interested in integer solutions) and it can be solved once the parameters $a$, $b$, $s$ and $d$ are known, but I fear it is not possible to give an explicit solution valid for all the cases, without knowing the parameters $a$, $b$, $s$ and $d$.
For example, if $d=1$, $s=2$, $a=2$ and $b=3$ there are no solutions.
If $d=1$, $s=3$, $a=41$ and $b=13$, there is only one solution with $n=226$ and $m=235$.
If $d=5$, $s=1$, $a=19$ and $b=6$ there are $3$ solutions, $(n,m)=(8,14)$ and $(63,74)$ and $(143,155)$ and so on.
A: thank you very much for your reply. Can I give you a better model and see what you think? This is actually the exact model. As I said I am self taught and have looked at what you have posted but will need the weekend to understand it better. This model I'm posting now is the exact problem I am looking at.
I have two sets A and B generated as
sequences, each of which is the sum of n terms of an arithmetic
series, I want to find all numbers that are in both sets (that
is, that occur as sums of both series. Both series increases by 2 for each move. 
Series A begins a movement of 6, then goes 8,10,12,14,16....
It's not important that both series start at the same time, time is irrelevant. 
Series B starts 6 feet behind series A. No matter what numbers I use series B will always start at some given integer behind A. In this model from this 6 foot position of B behind series A, B will move 35, 37, 39...
There will be a termination point for both neither runs to infinity. As you will see in this short example both series have a mark at the same spot, the 16 foot marker for series A, which the series sums to 66, and the the 37 foot marker for car B, where the series sums to 72-6=66.
 So this is the basic model. Sorry to repeat but I need to clarify in case I confused you with what I wrote previously. Is there a way to calculate where the series will have markers meet, in instances where different series do indeed have markers meet, without having to sum each segment of each series till they give out the same number?
