S1,S2,S3是等比數(shù)列,設(shè)公比為q
S2/S1=q
S3/S2=q
又Sn為等差數(shù)列的和,設(shè)公差為d
Sn=a1n+n(n-1)d/2
則S1=a1
S2=2a1+d
S3=3a1+3d
S2/S1=(2a1+d)/a1
=2+d/a1=q①
S3/S2=(3a1+3d)/(2a1+d)
=(3+3d/a1)/(2+d/a1)
=(3+3d/a1)/q
=q
則
3+3d/a1=q^2②
①②聯(lián)立得
q-2=(q^2-3)/3
3q-6=q^2-3
q^2-3q+3=0
無解,我估計(jì)你是不是把S1.S2.S3的順序搞錯(cuò)了?
2\
S2=4,可得4a1+6d
然后根據(jù)求出的q=a1/d+3
可求出a1和d
進(jìn)而求出Sn搞錯(cuò)了，是S4思路給了把S4代進(jìn)去S4=4a1+6dS4/S2=(4a1+6d)/(2a1+d)=2(2+3d/a1)/(2+d/a1)=2(2+3d/a1)/q=q2(2+3d/a1)=q^2聯(lián)立得[(q^2)/2-2]/3=q-2q^2/2=3q-4q^2-6q+8=0(q-2)(q-4)=0公比為2或4若公比為2,則a1/d=2-2=0a1=0S4=4a1+6d=4d=2/3Sn=n(n-1)/3若公比為4即a1/d=4-2=2且S4=4a1+6d=4兩式聯(lián)立4a1+3a1=4a1=5/7d=5/14Sn=(5/7)n+5n(n-1)/32再通分一下即可