7. (5 分) 已知 $\left\{a_{n}\right\}$ 是公差为 1 的等差数列, $S_{n}$ 为 $\left\{a_{n}\right\}$ 的前 $n$ 项和, 若 $S_{8}=4 S_{4}$, 则 $\mathrm{a}_{10}=(\quad)$
Gaokao · Math · previous-year question
- A.$\frac{17}{2}$
- B.$\frac{19}{2}$correct
- C.10
- D.12
Answer
B. $\frac{19}{2}$
Explanation
解: $\because\left\{a_{n}\right\}$ 是公差为 1 的等差数列, $S_{8}=4 S_{4}$, $\therefore 8 \mathrm{a}_{1}+\frac{8 \times 7}{2} \times 1=4 \times\left(4 \mathrm{a}_{1}+\frac{4 \times 3}{2}\right)$, 解得 $a_{1}=\frac{1}{2}$. 则 $\mathrm{a}_{10}=\frac{1}{2}+9 \times 1=\frac{19}{2}$. 故选: B.
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