Given two arithmetic sequences {an}, {BN}, the sum of the first n terms is Sn, TN respectively, and Sn / TN = 7n + 2 / N + 3, then A7 / B8= The answer is 31 / 6

Given two arithmetic sequences {an}, {BN}, the sum of the first n terms is Sn, TN respectively, and Sn / TN = 7n + 2 / N + 3, then A7 / B8= The answer is 31 / 6

Sn = (n / 2) (a1 + an), TN = (n / 2) (B1 + BN), let an tolerance be D1, BN tolerance be d2sn / TN = (a1 + an) / (B1 + BN) = (ND1 + a1-d1) / (ND2 + b1-d2) = (7n + 2) / (n + 3) let D2 = m, m ≠ 0, then D1 = 7m, a1-d1 = 2m, b1-d2 = 3M, then A1 = 9m, B1 = 4ma7 = a1 + 6d1 = 9m + 42m =

It is known that the sum of the first n terms of the sequence an is Sn, satisfying 8Sn = (an) ^ 2 + 4An + 3 (n ∈ n +) and A1, A2, a7 are the first three terms of the proportional sequence BN in turn
The general formula for finding an and BN

When 8s1 = 8A1 = 8A1 = A1, and when 8s1 = 8aa1 = 8A1 = A1 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\(n-1) & #178; - 4A

Given that in the equal ratio sequence an, a3a11 = 4a7, BN is the equal difference sequence, and B7 = A7, then what is the value of B5 + B9

a3a11=a7a7
So A7 = 4
b5+b9=2b7=8

It is known that the sequence {an} is an equal ratio sequence, Sn is the sum of the first n terms, A1, a7 and A4 are equal difference sequences, and it is proved that 2s3, S6, s12-s6 are equal ratio sequences?

A1, a7 and A4 are equal difference sequence
2a7=a1+a4
2a1*q^6=a1(1+q^3)
2q^6-q^3-1=0
(2q^3+1)(q^3-1)=0
q^3=1,q=1
Or: Q ^ 3 = - 1 / 2
If q = 1, then 2s3 = S6 = s12-s6 = 6A1, in an equal ratio sequence
If: Q ^ 3 = - 1 / 2
2s3=2a1(q^3-1)/(q-1)
s6=a1(q^6-1)/(q-1)
s12-s6=a1(q^12-q^6)/(q-1)=a1q^6(q^6-1)/(q-1)
s6/2s3=(q^6-1)/2(q^3-1)=(q^3+1)/2=1/4
(s12-s6)/s6=q^6=(q^3)^2=1/4
∴s6/2s3=(s12-s6)/s6
The sequence of 2s3, S6, s12-s6 is equal