等差数列An,Bn的前n项和分别为Sn,Tn且Sn除以Tn等于(3n-1)/(2n+3),求a8除以b8的值a8/b8=2*a8 / 2*b8=(a1+a15)/(b1+b15)=S15/T15=29/33

a8/b8=2*a8 / 2*b8=(a1+a15)/(b1+b15)=S15/T15=29/33不对吧s15=a1+a2+a3+a4+a5+a6+a7+a8+a9+a10+a11+a12+a13+a14+a15

an= a1+(n-1)d1
Sn=a1+a2+..+an = n[2a1+(n-1)d1]/2
bn= b1+(n-1)d2
Tn=b1+b2+..+bn=n[2b1+(n-1)d2]/2
Sn/Tn=[2a1+(n-1)d1]/[2b1+(n-1)d2]= (3n-1)/(2n+3)
put n=15
[2a1+14d1]/[2b1+14d2]=44/33
(a1+7d1)/(b1+7d2)=4/3
a8/b8=4/3