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9. 记 $S_{n}$ 为等差数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和. 已知 $S_{4}=0, a_{5}=5$, 则

Gaokao · Math · previous-year question

  1. A.$a_{n}=2 n-5$correct
  2. B.$a_{n}=3 n-10$
  3. C.$S_{n}=2 n^{2}-8 n$
  4. D.$S_{n}=\frac{1}{2} n^{2}-2 n$

Answer

A. $a_{n}=2 n-5$

Explanation

【详解】由题知, $\left\{\begin{array}{l}S_{4}=4 a_{1}+\frac{d}{2} \times 4 \times 3=0 \\ a_{5}=a_{1}+4 d=5\end{array}\right.$, 解得 $\left\{\begin{array}{l}a_{1}=-3 \\ d=2\end{array}, \therefore a_{n}=2 n-5\right.$, 故选 A.

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