已知等差数列an,bn的前n项和为Sn,Tn,Sn/Tn=3n-1/2n-3,求a7/b3,(a2+a5+a17+a22

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已知等差数列an,bn的前n项和为Sn,Tn,Sn/Tn=3n-1/2n-3,求a7/b3,(a2+a5+a17+a22)/(b8+b10+b12+b16)

由Sn/Tn=3n-1/2n-3,故设Sn=kn(3n-1),Tn=kn(2n-3)
a7=S13/13,b3=T5/5
a7/b3=5S13/13T5=[5*13k(13*3-1)]/[13*5k(2*5-3)]=38/7
A2+A5+A17+A22=(A1+d)+(A1+4d)+(A1+16d)+(A1+21d)=4A1+42d=2(2A1+21d)
同理B8+B10+B12+B16=4B1+42d'=2(2B1+21d')
S22=22A1+(1/2)×22×21d=22A1+231d=11(2A1+21d)
同理T22=11(2B1+21d')
(A2+A5+A17+A22)/(B8+B10+B12+B16)
=[2(2A1+21d)]/[2(2B1+21d')]
=[11(2A1+21d)]/[11(2B1+21d')]
=S22/T22=(3*22-1)/(2*22-3)=65/41

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