an=a+(n-1)c，bn=b+(n-1)d, Sn=a1+..+an=na+c(n-1)n/2 Tn=b1+..+bn=nb+d(n-1)n/2 (Sn)/(Tn)=(na+c(n-1)n/2)(nb+d(n-1)n/2)= =(c/d)(n+2a/c-1)/(n+2b/d-1)=(3/2)(n+5/3)/(n-3/2) 得c/d=3/2，2a/c-1=5/3，2b/d-1=-3/2， a/c=4/3，b/d=-1/4， a6/b6=（a+5c）/（b+5d）=(c/d)*(5+a/c)/(5+b/d）= =（3/2）*(5+4/3)/(5-1/4）=22/19。
答：设数列{an}首项为a1,公差为d， 所以b1=(1/2)^a1,b2=(1/2)^(a1+d),b3=(1/2)^(a1+2d), 所以(1/2)^(3a1+...详情>>
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