高中数学选择性必修 第二册（381题）

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知识点：第四章 数列

参考答案：设 \(\left\{ {{a_n}} \right\}\) 的公差为\(d\)，\(\left\{ {{b_n}} \right\}\) 的公比为\(q\)，\({b_1} = 2\)，\({a_1} = 1\)，

由题意可得：\(\left\{ {\begin{array}{*{20}{l}} {2q = 1 + d + 1} \\ {2{q^2} = 1 + 3d + 1} \end{array}} \right.\)，整理可得：\({q^2} - 3q + 2 = 0\)，

解得：\(\left\{ {\begin{array}{*{20}{l}} {q = 2} \\ {d = 2} \end{array}} \right.\) 或 \(\left\{ {\begin{array}{*{20}{l}} {q = 1} \\ {d = 0} \end{array}} \right.\)（舍）

所以 \({a_n} = 1 + \left( {n - 1} \right) \times 2 = 2n - 1\)，\({b_n} = 2 \cdot {2^{n - 1}} = {2^n}\)；