设等差数列\(\{a _{n} \}\)的前\(n\)项和为\(S _{n}\)，且\(5S _{5} =S _{10}\)，\(a _{4} =2a _{6} +20\)．
\((1)\)求数列\(\{a _{n} \}\)的通项公式；
\((2)\)若数列\(\{b _{n} \}\)满足\( \dfrac {b_{1}}{a_{1}} + \dfrac {b_{2}}{a_{2}} +…+ \dfrac {b_{n}}{a_{n}} = \dfrac {1}{2^{n}} -1\)，\(n∈N*\)，证明：\(b _{n} \leqslant \dfrac {5}{8}\)．