Question

设{an}是等差数列,bn=1/2^an,已知b1+b2+b3=21/8,b1*b2*b3=1/8,求等差数列的通项an

Answer 1

设bn的公比为q,首项为b

b+bq+bq^2=21/8 b^3q^3=1/8

所以bq=1/2 解得 b=1/8,q=4

b=2,q=1/4

当b=1/8,q=4,则d=-2,a1=3,an=5-2n

当b=2,q=1/4 则d=2,a1=-1 an=2n-3