解:设等差数列的首项为a1,等差是d。a2=a1+d,a4=a1+3d,即a1+3d=8,即a1=8-3d ①
又由于a1,a3,a7为等比数列,其中a3=a1+2d,a7=a1+6d,有:
(a1+2d)/a1=(a1+6d)/(a1+2d)
将①式代入上式: (8-3d+2d)/(8-3d)=(8-3d+6d)/(8-3d+2d),即(8-d)/(8-3d)=(8+3d)/(8-d)
64-16d+d^2=64-9d^2,即10d^2-16d=0
2d(5d-8)=0,所以d=8/5,a1=8-(24/5)=16/5
所以,an=a1+(n-1)d=(16/5)+(8/5)(n-1)
=(8/5)(n+1)
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