Question

已知an是递增的等差数列,且满足a2+a4=8 a1,a3,a7成等比数列？

求通项公式

Answer 1

解:设等差数列的首项为a1,等差是d。a2=a1+d，a4=a1+3d，即a1+3d=8，即a1=8-3d ①
又由于a1,a3,a7为等比数列，其中a3=a1+2d,a7=a1+6d，有:
(a1+2d)/a1=(a1+6d)/(a1+2d)
将①式代入上式: (8-3d+2d)/(8-3d)=(8-3d+6d)/(8-3d+2d)，即(8-d)/(8-3d)=(8+3d)/(8-d)
64-16d+d^2=64-9d^2，即10d^2-16d=0
2d(5d-8)=0，所以d=8/5，a1=8-(24/5)=16/5
所以，an=a1+(n-1)d=(16/5)+(8/5)(n-1)
=(8/5)(n+1)

Answer 2

a2+a4=2a3=8
a3=4，a1a7=(a3)²=64
a1(a1+6d)=64，又a3=a1+2d=4
联立求a1与d

Answer 3

手写详解！