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在等差数列 \(\{ {a_n}\} \) 中,已知 \({a_1} + {a_2} + {a_3} = 18\) 且 \({a_4} + {a_5} + {a_6} = 54\).
求 \(\{ {a_n}\} \) 的通项公式;
参考答案:解:由题意,设等差数列 \(\{ {a_n}\} \) 的公差为 \(d\),则 \(3{a_1} + 3d = 18\),\(3{a_1} + 12d = 54\), 解得 \({a_1} = 2\),\(d = 4\) \(\therefore \)\({a_n} = 2 + 4(n - 1) = 4n - 2\),\(n \in {N^*}\);
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