已知等比数列｛an}为a1+a2+a3+a4=15/8,a2*a3=-9/8,则1/a1+1/a2+1/a3+1/a4=

假设q=1,则A2=A3,A2×A3=A2^2>0,不符合题意
q≠1
A1+A2+A3+A4=A1×(1-q^4)/(1-q)
{1/An}是首项为1/A1,公比为1/q的等比数列
1/A1+1/A2+1/A3+1/A4=(1/A1)×(1-1/q^4)/(1-1/q)
两式相除
A1×(1-q^4)/(1-q)/[(1/A1)×(1-1/q^4)/(1-1/q)]
=A1^2×q^4/q×(1-q^4)/(1-q)/[(q^4-1)/(q-1)]
=A1^2×q^3
=A2×A3=-9/8
1/A1+1/A2+1/A3+1/A4
=(A1+A2+A3+A4)/(A2×A3)
=(15/8)/(-9/8)
=-5/3