Question

已知等差数列an中,a2＝4,a4是a2与a8的等比中项

Answer 1

设an=a1+(n-1)d

a2=a1+d=4

a4=a1+3d

a8=a1+7d

a4是a2与a8的等比中项

那么 a4^2=a2×a8

即

(a1+3d)^2=(a1+d)(a1+7d)=4(a1+7d)

即(a1+d+2d)^2=4(a1+d+6d) ,a1+d=4

(4+2d)^2=4(4+6d)

解之得 d=0或2

d=0 ,a1=4 ,an=4 ,即为常数数列

d=2,a1=2,an=2n

所以 数列an=4 或an=2n