Question

设数列{an}是等差数列，bn=(1/2)的an次方，又b1+b2+b3=21/8,b1b2b3

设数列{an}是等差数列，bn=(1/2)的an次方，又b1+b2+b3=21/8,b1b2b3=1/8,求通项an

Answer 1

bn为等比数列公比为q
b1=b2/q b3=b2q
带入b1b2b3=1/8得b2=1/2
由b1+b2+b3=21/8得q=1/4或4
q=1/4时 bn=2(1/4)^(n-1)=1/2^(2n-3) an=2n-3
q=4时，bn=1/8(4)^(n-1)=2^(2n-5) an=5-2n