求帮忙解答一道等比数列的题!!不难,要有过程

2025-05-14 05:46:44
推荐回答(1个)
回答1:

a2*a4=(a3)^2=1,得a3=1,S3=a1+a2+a3=a3(1/q^2+1/q+1)=7,所以1/q^2+1/q+1=7
,6q^2-q-1=0,(2q-1)(3q+1)=0,解得q=1/2,a1=a3/q^2=4,a4=a3*q=1/2,a5=a4*q=1/4,
S5=S3+a4+a5=7+1/2+1/4=31/4