Err not really.
I am not familiar with the tables, or figures, but if I understand it, you are talking the R + X values for each cable section, converting each to a Z value (ignoring the phase angle) and adding all the Z's.
Technically this is incorrect, if you are adding phasor values (vectors) then you must add including the phase angle - imagine one part of the cable is pure inductance R=0, X=1 ohm (Z=1), and one is pure capacitance again R=0, X=1 ohm (Z=1).
Using your method this would give a total of 2 ohms, however if you remember the phase angle, this is 90degree leading for 1 and 90degree lagging for the other, giving a total answer of 0 ohms.
In practice all the cables are likely to have a similar x/r ratio - so won't actually be radically different like my extreme example.
Add all the resistances and all the reactances first, then convert to Z
As an aside, you appear to have a shower load ~20A at 120V at the end of a cable 137 metres long - is this realistic?
We would usually have the dist board much closer so it is supplied with larger cables, in your case, you are likely to drop ~18V or so. (obviously I don't know what the actual current is, but even so it looks a bit wrong)
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Here is how I would use the figures:
Note, I think the layout is important as it helps, also to signify the reactive component I used the prefix j in the layout.
I haven't checked the values are correct from the table as I wouldn't know which ones are correct or not, so I just used what I understand are the correct R & X from your calcs.
Btw in the UK for normal circuits - basically anything below ~25mm^2 we only use the resistance, as the X/R ratio is ~10 and sqrt(10^2 + 1^2) is pretty much 10.
In this case, we would either use the standard, or the on-site guide, this shows for say a 2.5mm^2 cable with the same size earth/cpc a resistance of 14.82 ohm per kilometre so 137M would be 2 ohm amd then use that (obviously this is different as your choice of standard cable is slightly bigger than our standard cable here)