2023-2024衡水金卷先享题摸底卷 新教材A数学考试卷

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2023-2024衡水金卷先享题摸底卷 新教材A数学考试卷试卷答案

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分析设等差数列的公差为d，由题意表示出奇数项之和、奇偶项之和，利用等差数列的性质化简求出d的值．

解答解：由题意设等差数列的公差为d，
S_奇=a₁+a₃+a₅+…+a₁₉=15，①
S_偶=a₂+a₄+a₆+…+a₂₀=30，②
②-①得，10d=15，则d=$\frac{3}{2}$，
故选：B．

点评本题考查等差数列性质的灵活应用，以及整体思想，属于中档题．