Question

高中数学，求等差数列。

Answer 1

Sn=(a1+an) n/2，Tn=(b1+bn) n/2
Sn/Tn=(a1+an)/(b1+bn)=(7n+1)/(4n+27)=(2a1-d+nd)/(2b1-D+nD)
则d/D=7/4，(2a1-d)/(2b1-D)=1/27
a1/b1=S1/T1=8/31
所以 a1=(8/31)b1，d=(7/4)D
[2(8/31)b1-(7/4)D]/(2b1-D)=1/27
解出b1/D=31/8
设D=8t，则b1=31t，a1=(8/31)b1=8t，d=7/4 D=14t
a7/b7=(a1+6d)/(b1+6D)
=(8t+6*14t)/(31t+48t)
=92/79

Answer 2

an=a1+(n-1)d1
bn=b1+(n-1)d2
Sn/Tn = (7n+1)/(4n+27)
[2a1+(n-1)d1]/[2b1+(n-1)d2] =(7n+1)/(4n+27)
n=13
[2a1+12d1]/[2b1+12d2] =(91+1)/(52+27)
(a1+6d1)/(b1+6d2) = 92/79
a7/b7 =92/79