$\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{a}_{2}}}{{{a}_{3}}}=\frac{{{a}_{3}}}{{{a}_{4}}}=.........=\frac{{{a}_{2017}}}{{{a}_{2018}}}=\frac{1}{k}$. Suy ra:

${{a}_{2}}=k{{a}_{1}}$

${{a}_{3}}=k{{a}_{2}}={{k}^{2}}{{a}_{1}}$

........................

${{a}_{2018}}=k{{a}_{2017}}={{k}^{2017}}{{a}_{1}}$

$\Rightarrow$ $\frac{{{a}_{1}}}{{{a}_{2018}}}=\frac{{{a}_{1}}}{{{k}^{2017}}{{a}_{1}}}=\frac{1}{{{k}^{2017}}}=-{{5}^{2017}}$$\Rightarrow k=\frac{-1}{5}$ .

Theo tính chất của dãy tỉ số bằng nhau ta có

$\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{a}_{2}}}{{{a}_{3}}}=\frac{{{a}_{3}}}{{{a}_{4}}}=.........=\frac{{{a}_{2017}}}{{{a}_{2018}}}=S=\frac{{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+.........+{{a}_{2017}}}{{{a}_{2}}+{{a}_{3}}+{{a}_{4}}+.........+{{a}_{2018}}}=-5$