已知在等差数列\(\left\{ {{a_n}} \right\}\)中，\({a_4} + {a_8} = 20\),\({a_7} = 12\)，则\({a_9}\)=（       ）

设公差为\(d\)，

则\(\left\{ {\begin{array}{*{20}{l}}
{{a_1} + 3d + {a_1} + 7d = 20} \\\
{{a_1} + 6d = 12}
\end{array}} \right.\)，解得\(\left\{ {\begin{array}{*{20}{l}}
{{a_1} = 0} \\\
{d = 2}
\end{array}} \right.\)，

所以\({a_9} = {a_1} + 8d = 16\).

故选：D.

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