已知两个等差数列{an},{bn},其前n项和分别为Sn,Tn,且Sn/Tn=7n+2/n+3,则a7/b8=
问题描述：
已知两个等差数列{an},{bn},其前n项和分别为Sn,Tn,且Sn/Tn=7n+2/n+3,则a7/b8=
答案是31/6
答
sn=（n/2）（a1+an）,Tn=（n/2）（b1+bn）,设an公差为d1,bn公差为d2Sn/Tn=（a1+an）/（b1+bn）=（nd1+a1-d1）/（nd2+b1-d2）=（7n+2）/（n+3）令d2=m,m≠0,则d1=7m,a1-d1=2m,b1-d2=3m得a1=9m,b1=4ma7=a1+6d1=9m+42m=...答案是31/6思路对的，Sn/Tn=（a1+an）/（b1+bn）=（nd1+2a1-d1）/（nd2+2b1-d2）=（7n+2）/（n+3）令d2=m，m≠0，则d1=7m，2a1-d1=2m，2b1-d2=3m得a1=（9/2）m，b1=2ma7=（9/2）m+42m=（93/2）mb8=2m+7m=9ma7/b8=93/18=31/6