A system of sums

The sequences shown have an alternating geometric ratio. For the first equation, we have $x$ first multiplied, then $y$ , and then $x$ again and so on. On the other hand, it is the other way around on the second equation. That is, $y$ first, and then $x$ , and so on.

Based on the given equations above, we can group the terms of each sequence in pairs, such that it will form a geometric sequence in one geometric ratio $xy$ .

$\large (1 + x) + (1+x)xy + (1+x)(xy)^{2} + (1+x)(xy)^3 + ... = \frac{21}{5} = \frac{1+x}{1-xy}$

Similarly, we can do this for the other equation, and get

$\large \frac {1+y}{1-xy} = 4$

So now we condensed the system into a form which can be solved by basic algebraic manipulation. This will lead us to

$5 + 5x = 21 - 21xy$ $1 + y = 4 - 4xy$

Under the condition that $x$ , $y$ , and $xy$ should be less than 1, we find that $x = \frac{4}{5}$ , and $y = \frac{5}{7}$ .

Now to find $z$ , we need to simplify first the given expression. In this case, the geometric ratio recurs every three terms, so the sum is found as

$\large \frac { 1 + \frac{x}{2} + \frac{xy}{4}}{1 - \frac {xyz}{8}} = \frac {216}{133}$

Simplifying, we get

$\large \frac { 1 + \frac {2}{5} + \frac {1}{7}}{1 - \frac{z}{14} } = \frac {216}{133}$ $\large \frac {108}{5(14-z) } = \frac {216}{133}$

and voila! $\large z = \frac {7}{10} = \boxed {0.7}$

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x=4/5; y=5/7; z=7/10