设数列 $ {a_n} $ 的公比为 $ q\left(q\neq 0,q\neq 1\right) $，
由 $ a_5,a_3,a_4 $ 成等差数列，得 $ 2a_3=a_5+a_4 $，即\[ 2a_1q^2=a_1q^4+a_1q^3 ,\]由 $ a_1\neq 0 ,q\neq 0$，得 $ q^2+q-2=0 $，解得\[ q_1=-2,q_2=1\left(舍去\right), \]所以\[ q=-2. \]

证法一：对任意 $ k\in {\mathbb{N}}_+ $，\[ \begin{split}S_{k+2}+S_{k+1}-2S_k &=\left(S_{k+2}-S_k\right)+\left(S_{k+1}-S_k\right)\\&=a_{k+1}+a_{k+2}+a_{k+1}\\&=2a_{k+1}+a_{k+1}\cdot \left(-2\right)\\&=0, \end{split}\]所以，对任意 $ k\in {\mathbb{N}}_+,S_{k+2},S_k,S_{k+1} $ 成等差数列．
证法二：对任意 $ k\in {\mathbb{N}}_+ $，\[ \begin{split}2S_k&={\dfrac{2a_1\left(1-q^k\right)}{1-q}},\\S_{k+2}+S_{k+1}& ={\dfrac{a_1\left(1-q^{k+2}\right)}{1-q}}+{\dfrac{a_1\left(1-q^{k+1}\right)}{1-q}}\\&={\dfrac{a_1\left(2-q^{k+2}-q^{k+1}\right)}{1-q}},\\2S_k-\left(S_{k+2}+S_{k+1}\right) &={\dfrac{2a_1\left(1-q^k\right)}{1-q}}-{\dfrac{a_1\left(2-q^{k+2}-q^{k+1}\right)}{1-q}}\\&={\dfrac{a_1}{1-q}}\left[2\left(1-q^k\right)-\left(2-q^{k+2}-q^{k+1}\right)\right]\\&={\dfrac{a_1q^k}{1-q}}\left(q^2+q-2\right)\\&=0, \end{split}\]所以，对任意 $ k\in {\mathbb{N}}_+,S_{k+2},S_k,S_{k+1} $ 成等差数列．