已知等差数列\(\left\{ {{a_n}} \right\}\)满足\({a_1} + {a_3} + {a_5} = 12\)，\({a_{10}} + {a_{11}} + {a_{12}} = 24\)，则\(\left\{ {{a_n}} \right\}\)的前\(13\)项的和为（       ）

设等差数列\(\left\{ {{a_n}} \right\}\)公差为\(d\)，

\(\because {a_1} + {a_3} + {a_5} = 12\)，\({a_{10}} + {a_{11}} + {a_{12}} = 24\)，\(\therefore {a_{10}} + {a_{11}} + {a_{12}} - \left( {{a_1} + {a_3} + {a_5}} \right) = 24d = 12\)，

解得：\(d = \frac{1}{2}\)，\(\therefore {a_1} + {a_3} + {a_5} = 3{a_1} + 6d = 3{a_1} + 3 = 12\)，解得：\({a_1} = 3\)，

\(\therefore \left\{ {{a_n}} \right\}\)的前\(13\)项的和为\(13{a_1} + \frac{{13 \times 12}}{2}d = 39 + \frac{{13 \times 12}}{4} = 78\).

故选：C.

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