题目内容
| 1 |
| 8 |
| 3 |
| 16 |
| 1 |
| 2 |
| 1 |
| 2 |
分析:根据题设条件每行数都成等差数列,可先求出a41,根据每列数都成等比数列,且所有公比都相同,可求公比q,进而可求a11.
解答:解:由题意,∵每行数都成等差数列,a42=
,a43=
∴a41=
每列数都成等比数列,且所有公比都相同,设每列数的公比为q
∵a12=1,a42=
,
∴q=
∵a41=a11×q3,
∴a11=
故答案为:
| 1 |
| 8 |
| 3 |
| 16 |
∴a41=
| 1 |
| 16 |
每列数都成等比数列,且所有公比都相同,设每列数的公比为q
∵a12=1,a42=
| 1 |
| 8 |
∴q=
| 1 |
| 2 |
∵a41=a11×q3,
∴a11=
| 1 |
| 2 |
故答案为:
| 1 |
| 2 |
点评:本题考查数列的运用及等差数列的性质与等比数列的性质,认真审题,理解领会题意是解题的关键
练习册系列答案
相关题目