A = \(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{17\cdot20}=\dfrac{1}{3}\cdot\left(\dfrac{3}{2}-\dfrac{3}{5}+\dfrac{3}{5}-\dfrac{3}{8}+...+\dfrac{3}{17}-\dfrac{3}{20}\right)=\dfrac{1}{3}\cdot\left(\dfrac{3}{2}-\dfrac{3}{20}\right)=\dfrac{1}{3}\cdot\dfrac{27}{20}=\dfrac{9}{20}\)