设{an}公差为d,{bn}公差为d'.
Sn/Tn=[na1+n(n-1)d/2]/[nb1+n(n-1)d'/2]
=[dn+(2a1-d)]/[d'n+(2b1-d')]
=(2n+1)/(3n+2)
令d=2t,则2a1-d=t,d'=3t,2b1-d'=2t
解得
a1=(3/2)t d=2t b1=(5/2)t d'=3t
a12/b15=(a1+11d)/(b1+14d')
=[(3/2)t +11×(2t)]/[(5/2)t+14×(3t)]
=(3/2 +22)/(5/2 +42)
=37/89