(1)等差数列an的公差为d1等差数列bn的公差为d2Sn/Tn=(7n+2)/(n+3)[n(a1+an)/2]/[n(b1+bn)/2]=(7n+2)/(n+3) (a1+a1+(n-1)*d1)/(b1+b1+(n-1)*d2)=(7n+2)/(n+3) (2a1-d+n*d1)/(2b1-d2+n*d2)=(7n+2)/(n+3) 于是分子分母比较相应系数2a1-d1=2k,d1=7k,2b1-d2=3k, d2=k解得a1=9k/2,d1=7k,b1=2k,d2=k即an=9k/2+(n-1)*7k)=7kn-5k/2 bn=2k+(n-1)*k)=kn+k故an/bn=(7kn-5k/2)/(kn+k)=(14n-5)/(2n+2)(2)an/bn=(14n-5)/(2n+2)(a1+(n-1)*d1)/(b1+(n-1)*d2)=(14n-5)/(2n+2)(a1-d1+n*d1)/(b1-d2+n*d2)=(14n-5)/(2n+2)于是分子分母比较相应系数a1-d1=-5k,d1=14k,b1-d2=2k, d2=2k解得a1=9k,d1=14k,b1=4k,d2=2k即Sn=n*a1+n(n-1)*d1/2=9kn+7kn(n-1)Tn=n*b1+n(n-1)*d2/2=4kn+kn(n-1)故Sn/Tn=[9kn+7kn(n-1)]/[4kn+kn(n-1)]=[9+7(n-1)]/[4+(n-1)]=(7n+2)/(n+3)
