\(Q=\dfrac{3^2}{2.5}+\dfrac{3^2}{5.8}+...+\dfrac{3^2}{99.102}\\ =3\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{99.102}\right)\\ =3\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{99}-\dfrac{1}{102}\right)\\ =3\left(\dfrac{1}{2}-\dfrac{1}{102}\right)\\ =3.\dfrac{25}{51}\\ =\dfrac{25}{17}\)