Question

已知等比数列{an}的各项均为正数,且2a1+ 3a2=1,a3^2=9 a2a6.求数列的通项公式；

Answer 1

设正项等比数列{an}公比为q(q>0)

由a3^2=9 a2a6 =>(a3)^2=9(a3/q)*(a3*q^3)

=>(a3)^2=9(a3)^2*q^2

=> q=1/3

又2a1+ 3a2=1 =>2a1+3a1*q=1=>a1=1/3

通项公式an=a1*q^(n-1）=(1/3)^n