It is known that real number sequence an is equal ratio sequence, where A7 = 1, and A4, A5 + 1, A6 are equal difference sequence. The general term formula of sequence an is obtained fast

It is known that real number sequence an is equal ratio sequence, where A7 = 1, and A4, A5 + 1, A6 are equal difference sequence. The general term formula of sequence an is obtained fast

AQ ^ 3 = 1 / Q ^ 3, AQ ^ 3, AQ ^ 5 = 1 / Q A4, a 5 + 1, and a 6 is the difference sequence 2 (a * q ^ 4 + 1) (a * q ^ 4 + 1) = a * q ^ 3 + A * q ^ 5 2A * q ^ 4 + 2 = a * q ^ 3 = 1 / Q ^ 2, AQ ^ 3 = 1 / Q ^ 3, AQ ^ 3, AQ ^ 3 = 1 / Q ^ 3, a / Q ^ 3, AQ ^ 3, AQ ^ 3, AQ ^ 3, AQ ^ 3 = 1, AQ ^ 2q ^ 2q ^ 2q ^ 2q ^ 2q ^ 2q ^ 2q ^ 2 (2q-1 (2q-1) + (2q-2q-2q-1) + (2q-q-1)) = 0 (Q ^ 2 (Q ^ 2) (Q ^ 2) (2q ^ 2) (2q ^ 2) (2q-1) (2q-1) (2q-1) (2q-2 ^ 6 = 64, so the general term an = 64 * (1 / 2) ^ (n-1)

If an = 1 / 2, find n (2) let the sum of the first n terms of the sequence be Sn, find S8

If the solution is as follows, (1) let the common ratio of the equal ratio sequence {an} be q, then an = a1qn-1, A3 + A6 = 36, A4 + A7 = 18, q = 0. A3 + A6 = 36 = a1q ^ 2 (1 + Q ^ 3), then A3 = 32, then A1 = 128, let an = 1 / 2 = a1q ^ (n-1) solve N = 9, S8 = a1 + A2 + ---- + A8 = (use the formula) = 255

Let the common ratio | Q | of the equal ratio sequence an be greater than 1, and let A3 = 2, S4 = 5s2 find A5 + A7

When | Q | > 1, Sn = A1 (1-Q ^ n) / (1-Q)
s4=5s2 ==>1-q^4=5(1-q^2)
Q ^ 2 = 4
A3 = a1q ^ 2 = 4A1 = 2, calculate A1 = 1 / 2
a5+a7=a1q^4+a1q^6=8+32=40

In the positive proportional sequence {an}, Sn is the sum of the first n terms, A3 = 2, S4 = 5s2, then A5=______ ．

From S4 = 5s2, we know that the common ratio Q ≠ 1 of the sequence, from A3 = 2, S4 = 5s2, we get a1q2 = 2, A1 (1 − Q4) 1 − q = 5A1 (1 − Q2) 1 − Q, simplify to 1 + Q2 = 5, ②, simultaneous solution to A1 = 12, q = 2, so A5 = a1q4 = 12 × 24 = 8, so the answer is: 8