由实数构成的等比数列 \(\left\{ {{a_n}} \right\}\) 的前\(n\)项和为 \({S_n}\) ， \({a_1} = 2\) ，且 \({a_2} - 4,{a_3},{a_4}\) 成等差数列，则 \({S_6} = \) （       ）

由题意知， \(2{a_3} = {a_2} - 4 + {a_4}\) ，

设 \(\left\{ {{a_n}} \right\}\) 的公比为 \(q\) ，则 \(\left\{ {\begin{array}{*{20}{c}}
{2{a_1}{q^2} = {a_1}q - 4 + {a_1}{q^3}} \\\
{{a_1} = 2}
\end{array}} \right.,\)

解得 \(q = 2\) ，则 \(\frac{{2\left( {1 - {2^6}} \right)}}{{1 - 2}} = 126\) .

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