a,\(\frac{-3}{1.3}+\frac{-3}{3.5}+....+\frac{-3}{97.99}\)
= -3.\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)\)
=\(\frac{-3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)\)
=\(\frac{-3}{2}\left(1-\frac{1}{99}\right)\)
=\(\frac{-3}{2}.\frac{98}{99}\)
=\(\frac{49}{-33}\)>\(\frac{49}{-20}\)