Question

设等比数列 {an}的公比q

Answer 1

等比数列an=a1*q^(n-1),Sn=a1(1-q^n)/(1-q)

∴a3=2=a1*q^(3-1)=a1*q^2

S4=5S2 => a1(1-q^4)/(1-q)=5*a1(1-q^2)/(1-q)

=> 1-q^4=5(1-q^2) => q^4-5q^2+4=0 解得q^2=1或q^2=4

∵q