\(A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)....\left(1-\frac{1}{100}\right)=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.......\frac{99}{100}=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}.....\frac{9.11}{10^2}=\frac{\left(1.2.3....9\right).\left(3.4.5....11\right)}{\left(2.3.4....10\right).\left(2.3.4....10\right)}=\frac{1.11}{10.2}=\frac{11}{20}\)