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4. (5 分) 记 $S_{n}$ 为等差数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和. 若 $a_{4}+a_{5}=24, S_{6}=48$, 则 $\left\{a_{n}\right\}$ 的公 差为 $(\quad)$

Gaokao · Math · previous-year question

  1. A.1
  2. B.2
  3. C.4correct
  4. D.8

Answer

C. 4

Explanation

解: $\because S_{n}$ 为等差数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和, $a_{4}+a_{5}=24, S_{6}=48$, $\therefore\left\{\begin{array}{l}a_{1}+3 d+a_{1}+4 d=24 \\ 6 a_{1}+\frac{6 \times 5}{2} d=48\end{array}\right.$, 解得 $\mathrm{a}_{1}=-2, \mathrm{~d}=4$, $\therefore\left\{a_{n}\right\}$ 的公差为 4 . 故选: C.

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