已知等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，\(S_{5}=-5\)，且\(a_{3}\)，\(a_{4}\)，\(a_{6}\)成等比数列．
\((1)\)求数列\(\{a_{n}\}\)的通项公式；
\((2)\)设\(b_{n}= \dfrac {1}{a_{2n+1}a_{2n+3}}(n∈N^{*})\)，求数列\(\{b_{n}\}\)的前\(n\)项和\(T_{n}\)．