A=$\frac{5}{1.3}$+$\frac{5}{3.5}$+...+$\frac{5}{99.101}$
=$\frac{5}{2}$ .($\frac{2}{1.3}$+$\frac{2}{3.5}$+...+$\frac{2}{99.101}$ )
=$\frac{5}{2}$ .($1-\frac{1}{3}$+$\frac{1}{3}$-$\frac{1}{5}$+...+$\frac{1}{99}$-$\frac{1}{101}$ )
=$\frac{5}{2}$ .($1-\frac{1}{101}$)
=$\frac{5}{2}$ .\frac{100}{101}$
=$\frac{250}{101}$