等差数列{an}{bn},Sn,Tn分别是前n项和,且Sn/Tn=(7n+1)/(n+3),则（a2+a5+a17+a22)/(b8+b10+b12+b16)=

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等差数列{an}{bn},Sn,Tn分别是前n项和,且Sn/Tn=(7n+1)/(n+3),则（a2+a5+a17+a22)/(b8+b10+b12+b16)=

等差数列{an}{bn},Sn,Tn分别是前n项和,且Sn/Tn=(7n+1)/(n+3),则（a2+a5+a17+a22)/(b8+b10+b12+b16)=
等差数列{an}{bn},Sn,Tn分别是前n项和,且Sn/Tn=(7n+1)/(n+3),则（a2+a5+a17+a22)/(b8+b10+b12+b16)=

等差数列{an}{bn},Sn,Tn分别是前n项和,且Sn/Tn=(7n+1)/(n+3),则（a2+a5+a17+a22)/(b8+b10+b12+b16)=
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
=(7×22+1)/(22+3)
=31/5

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