CONCEPT: The question applies the concept of Progressions.

SOLUTION:
 
Since x22-x21=y99-y100
It implies the common difference of the two A.Ps are the negatives of each other.
[As, the common difference of 2nd AP = - (y99 - y100) = - (x22 - x21) = - d]
Thus the series,
(x1+y1),(x2+y2),is a constant series with each term equal to 100 (x1+y1=100)


Thus the required sum is
 100*100=10,000