\(A=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{29.31}{30^2}=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right)...\left(29.31\right)}{\left(2.3.4...30\right)^2}=\frac{\left(1.2.3...29\right).\left(3.4.5...31\right)}{\left(2.3.4...30\right).\left(2.3.4...30\right)}=\frac{1.31}{30.2}=\frac{31}{60}\)