设Sn、Tn分别是等差数列an、bn的前n项和,Sn/Tn=（7n+2)/（n+3）,则a6/b5=?

这个就要比a6/b6复杂多了.

设{an}公差为d1,{bn}公差为d2.
Sn/Tn=[na1+n(n-1)d1/2]/[nb1+n(n-1)d2/2]
=[2a1+(n-1)d1]/[2b1+(n-1)d2]
=[d1n+(2a1-d1)]/[d2n+(2b1-d2)]
=(7n+2)/(n+3)
令d1=7t,则2a1-d1=2t,d2=t,2b1-d2=3t
解得a1=(9/2)td1=7tb1=2td2=t
a6/b5=(a1+5d1)/(b1+4d2)
=[(9/2)t+5×7t]/(2t+4t)
=(9/2 +35)/6
=79/12
你的老师给的答案是正确的.