等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，已知\(a_{1}=10\)，\(a_{2}\)为整数，且\(S_{n}\leqslant S_{4}\)．
\((1)\)求\(\{a_{n}\}\)的通项公式；
\((2)\)设\(b_{n}= \dfrac {1}{a_{n}a_{n+1}}\)，求数列\(\{b_{n}\}\)的前\(n\)项和\(T_{n}\)．