What Would This Even Look Like?

Let $S$ denote the sum of all of the terms in the sequence.

By the given information, we know that

$x_{a}+x_{b}=S-x_{a}-x_{b}-a-b \\ 2x_{a}+2x_{b}=S-a-b$

Note that there are $101 \choose 2$ ways we can pair up two numbers in this sequence. Also note that there are 100 ways to pair $a_{0}$ ; once with $x_{1}$ , once with $x_{2}$ , and so on. This holds for any $x_{p}, a$ and $b$ .

This means if we sum up our above equation for all possible combinations of two terms of our sequence, we will get

$200(x_{0}+x_{1}+x_{2}+...+x_{100})={101 \choose 2} S - 100\sum_{k=0}^{100}k$

Here, we can simplify and get

$200S=5050S-(50)(100)(101) \\ 4850S=505000 \\ S=\frac{10100}{97} \\ S=104+\frac{12}{97}$

Now, we can set up the following system:

$2(x_{49}+x_{50})=104+\frac{12}{97}-49-50=5+\frac{12}{97} \\ 2(x_{49}+x_{51})=4+\frac{12}{97} \\ 2(x_{50}+x_{51})=3+\frac{12}{97}$

Subtract the second equation from the first to get

$2x_{50}-2x_{51}=1 \\ 2x_{50}+2x_{51}=3+\frac{12}{97}$

Add the two equations together to get

$4x_{50}=4+\frac{12}{97} \\ x_{50}=\frac{100}{97}$

Since 100 and 97 are co-prime, our answer is $97+100=\boxed{197}$

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As for the problem's title, the sequence actually turns out to be an arithmetic sequence, with each successive term decreasing by $\frac{1}{2}$

$\displaystyle Let ~S=\sum_{i=0}^{100} X_i.\\ X_a+X_b=S -(X_a+X_b) - (a+b) ~~~ \implies ~ X_a+X_b =\dfrac S 2 - \dfrac{a+b} 2\\ \implies ~ \color{#3D99F6}{X_i =\dfrac S 4 - \dfrac i 2 }, ~~0 \leq i \leq101.\\ \displaystyle ~S=\sum_{i=0}^{100} X_i=\sum_{i=0}^{100} \left ( \dfrac S 4 - \dfrac i 2 \right)=\dfrac{101}4 S- \dfrac{50*101} 2 . ~~~~ \therefore ~\dfrac{97} 4 S=25*101.\\ \implies ~\color{#3D99F6}{S= \dfrac{4*25*101} {97}}.\\ X_{50}=\dfrac S 4 - \dfrac {50} 2=\dfrac{25*101} {97} - 25=\dfrac{4*25}{97}=\dfrac m n.\\ m+n= \Huge ~~~~\color{#D61F06}{197}$