Cummulative blocks on a rope, tension problem

This isn't a joke, I really don't know if this is right or wrong. Please let me know if this is wrong so I can learn what I am not getting right.

1. The problem statement, all variables and given/known data
Three different sets of boxes are being pulled along frictionless surfaces attached to a rope that ends with the rightmost box, accelerating toward the left. All the boxes are identical, and the accelerations are as follows
1) 3a for one box (box is called a)
2) 2a for two boxes (from left to right, called b and c)
3) a for 3 boxes (from left to right, called d, e, and f)

Rank the magnitude of the tension in these ropes from largest to smallest.

2. Relevant equations
force is mass times acceleration.
magnitude means absolute value
tension=magnitude of the pulling force exerted by a string, cable, chain, or similar object on another object

This is my first point of confusion. What do they mean by magnitude of the tension? If tension is already a magnitude then what are they asking for?

3. The attempt at a solution

If the boxes are identical, then their masses should be the same. If that is correct, then it seems like the ropes would have the following tensions:
1) m(3a)
2) 2m(2a)
3) 3m (a)

It seems like the answer may be that 2 has the most and (1 and 3) are equal to each other but less than 2.

Is that right, can I just add the forces together like that and compare them? Or is this wrong?

 

Thank you both for your answers. It is so kind of you to spend time helping confused students. I truly appreciate what you are doing here.

I am attaching a scan of the question for clarification. I hope the question I asked earlier matches the picture. If not, could you please tell me how it does not match?

 

I'm sorry. I am so confused.

Trying again here--I editted the picture.
I have checked all the definitions again.


So in the gray box we have this rope. The acceleration of the 2block/1box unit is 2a. The mass of the 2 blocks is 2 each.

The force on the rope B is 4ma.
The mass of the second block is m. The acceleration of teh whole unit is 2a. So maybe the force on this second block is 2ma. Maybe the tension on the rope is C is opposite that of the block, but it can't be opposite because tension is always positive.
So that's wrong too.

Trying again--
The force on the rope B is 4ma.
The mass of the second block is m. The acceleration of the whole unit is 2a. The force on this second block may be 2ma. Since the answer is not 4ma and only positive values are allowed maybe I should add them. Could it be 6ma?

Thank you so much for helping me. I am so sorry to be asking such incredibly stupid questions.

 

trying again to add teh picture

 

Thanks again. So it was the first way. I was able to do several more problems given that information.

One of my big mistakes was not realizing that one box could be involved in multiple action-reaction pairs. I was trying to balance the boxes on both sides as well as the ropes which obviously is completely wrong.