Alan

Does this help?

Substitute the values of the coordinates given for P into the equation to get one equation for a and b; then substitute the coordinates given for Q into the equation to get another equation for a and b.  You now have two simultaneous equations for two unknowns, so you can solve for a and b.  (Tip: it is easier to make the substitutions u = 1/a and v = 1/b first; solve for u and v then set a = 1/u and b = 1/v).

Find the common ratio, r, given by the ratio of one term divided by the immediately preceeding term.  The infinite series sum is then a/(1-r) where a is the first term.

Here is one approach:

You could also look at it this way:

You can see that this will always end with 10/18, i.e. there will always be a remainder of 10.

$a+ar+ar^2+...=\frac{a}{1-r}$ as long as r<1

$a^3+a^3(r^3)+a^3(r^3)^2+...=\frac{a^3}{1-r^3}$  as long as r^3<1

So:  

$\frac{a}{1-r}=4$

and

$\frac{a^3}{1-r^3}=10$

Can you take it from here?

The other condition seems to be that you specify in advance the values of x, y and z.  It wasn't clear from your original question that that was what was intended!

There are an infinite number of solutions to the question you asked - see below.  Perhaps you meant to add another condition.

Like so:

Use $\lim_{n\rightarrow\infty}Summation$

I'll leave you to write the whole expression in LaTeX.

I couldn't see a neat way of doing this analytically, so I used a numerical approach:

As follows:

@hairyberry  You need to check $f(4\sqrt5)$.  It isn't 70.

"I wasn't really sure if you could separate the magnitude of z+w to the sum of two magnitudes."

You can't in general!

Any permutation starting 1, 5 , ...    4! of them 

Any permuttion starting 1, 6, ...       4! of them

Any permutation ending ..., 5, 1      4! of them

Any permutation ending ..., 6, 1      4! of them

Any permutation starting 5, 1, 6, ...  3! of them

Any permutation starting 6, 1, 5, ...  3! of them

etc. ...

The limit of the argument is zero, but that doesn't make the sum zero.

Use the following:

As follows:

@Padewolofoofy:  You forgot to divide by 2^2