记等差数列{an}的前n项和为Sn.设S3=12,且2a1,a2,a3+1成等比数列,求Sn

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记等差数列{an}的前n项和为Sn.设S3=12,且2a1,a2,a3+1成等比数列,求Sn记等差数列{an}的前n项和为Sn.设S3=12,且2a1,a2,a3+1成等比数列,求Sn记等差数列{an

记等差数列{an}的前n项和为Sn.设S3=12,且2a1,a2,a3+1成等比数列,求Sn
记等差数列{an}的前n项和为Sn.设S3=12,且2a1,a2,a3+1成等比数列,求Sn

记等差数列{an}的前n项和为Sn.设S3=12,且2a1,a2,a3+1成等比数列,求Sn
{an}是等差数列
S3=a1+a2+a3=3a2=12
a2=4
设公差为d
a1=4-d a3=4+d
2a1,a2,a3+1成等比数列
(a2)^2=2a1·(a3+1)
4^2=2(4-d)(4+d+1)
8=(4-d)(d+5)
8=20-d-d^2
d^2+d-12=0
(d+4)(d-3)=0
d=-4 或 d=3
若d=-4,则a1=8,an=a1+d(n-1)=8-4(n-1)=12-4n
Sn=(a1+an)n/2=(8+12-4n)n/2=-2n^2+10n
若d=3,则a1=1,an=a1+d(n-1)=1+3(n-1)=3n-2
Sn=(a1+an)n/2=(1+3n-2)n/2=(3/2)n^2-(1/2)n

S3=12
a1+a2+a3=12
a1+a3=2a2
a2=4
a2平方=2a1*(a3+1)
a3=8-a1
16=2a1(8-a1+1)
a1=正负1
舍掉-1
根据公式na1+n(n-1)d/2
得出sn

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