Question

若｛a_n｝既为等差数列，又为等比数列，求证a_n＝a₁，n∈N*．

 

Answer 1

答案：
解析：

【证明】 由已知a1＋d＝a1q                                                                                   ① a1＋2d＝a1q2                                                                                                                                                                                                                      ② 由①得a1(1－q)＝－d                                                                                              ③ 由②得a1(1－q2)＝－2d                                                                                                  ④ ∴a1(1－q2)＝2a1(1－q)                                                                                            ⑤ ∵｛an｝为等比数列，则a1≠0，q≠0 则由⑤，得1＋q＝2即q＝1 ∴an＝a1qn－1＝a1

提示：

只有非零的常数列既是等差数列，又是等比数列，也可利用等差中项和等比中项证明此结论．