**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