等差数列an的各项均为正数,a1=1前n项和为sn数列bn为等比数列b1=2且b2s2=16,b3s3=72.求an bn通项
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![等差数列an的各项均为正数,a1=1前n项和为sn数列bn为等比数列b1=2且b2s2=16,b3s3=72.求an bn通项](/uploads/image/z/7611597-45-7.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97an%E7%9A%84%E5%90%84%E9%A1%B9%E5%9D%87%E4%B8%BA%E6%AD%A3%E6%95%B0%2Ca1%3D1%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%E6%95%B0%E5%88%97bn%E4%B8%BA%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97b1%3D2%E4%B8%94b2s2%3D16%2Cb3s3%3D72.%E6%B1%82an+bn%E9%80%9A%E9%A1%B9)
等差数列an的各项均为正数,a1=1前n项和为sn数列bn为等比数列b1=2且b2s2=16,b3s3=72.求an bn通项
等差数列an的各项均为正数,a1=1前n项和为sn数列bn为等比数列b1=2且b2s2=16,b3s3=72.求an bn通项
等差数列an的各项均为正数,a1=1前n项和为sn数列bn为等比数列b1=2且b2s2=16,b3s3=72.求an bn通项
设{an}公差为d,{bn}公比为q.
等差数列{an}各项均为正,则公差d≥0
b2S2=16 b3S3=72
(b1q)×(2a1+d)=16
(b1q²)×(3a1+3d)=72
a1=1 b1=2代入,整理,得
q(d+2)=8 (1)
q²(d+1)=12 (2)
(1)²/(2)
(d+2)²/(d+1)=64/12
整理,得
3d²-4d-4=0
(3d+2)(d-2)=0
d=-2/3(