In the equal ratio sequence {an}, a1 + A7 = 65, A3 · A5 = 64, and an + 1 & lt; an (1) finds the general term of the sequence {an}. (2) finds the first five terms and s of the sequence {an}

In the equal ratio sequence {an}, a1 + A7 = 65, A3 · A5 = 64, and an + 1 & lt; an (1) finds the general term of the sequence {an}. (2) finds the first five terms and s of the sequence {an}

1、
A3 · A5 = 64, we get A1 & # 178; Q ^ 6 = 64, that is, Q ^ 6 = 64 / A1 & # 178; Q ^ 6 = 64 / A1 & # 178;
a1+a7=65
That is a1 + a1q ^ 6 = 65
a1+a1*64/a1²=65
a1²-65a1+64=0
The solution is A1 = 1 or A1 = 64
Because an + 1

In the arithmetic sequence an, D ≠ 0, BN is the arithmetic sequence with positive items, A1 = B1, A3 = B3, a7 = B5, A15 = BM, find M

Let the tolerance of arithmetic sequence an be D and the common ratio of arithmetic sequence BN be Q
a1=b1
a3=b3 => a1+2d=b3=a1*q^2
a7=b5 => a1+6d=b5=a1*q^4
b3²=b1*b5 => (a1+2d)²=a1*(a1+6d) =>a1=2d
So B3 = A3 = 4D, B5 = A7 = 8D
Because BN is an equal ratio sequence with positive terms, q = √ (B5 / B3) = √ 2
bn=b1*q^(n-1）=2d*√2^(n-1)=2d*2^[(n-1)/2]
a15=a1+14d=16d=2d*8=2d*2^3
bm=2d*2^[(m-1)/2]
From A15 = BM, (m-1) / 2 = 3
m=7

In the equal ratio sequence, a5-a1 = 15, a4-a2 = 6, find A3 and S6
Solution process ah, a little easier way, thank you

Let the ratio of the equal ratio sequence be Q
a1*(q^4-1)=15 (1)
a1*(q^3-q)=6 (2)
Formula (1) divided by formula (2): 2q ^ 2-5q + 2 = 0 (3)
Two solutions are obtained from (3) and then substituted into (1)
Then A3 = A1 * q ^ 2; S6 = A1 * (Q ^ 6-1) / (Q-1); calculate A3, S6

Let an satisfy a1 + 2A2 + 2 ^ 2A3 +... 2 ^ n-1an = n ^ 2 / 2
Can you help me to look at this problem again? I don't know how to do this series problem

Because: a1 + 2A2 + 2 ^ 2A3 +... 2 ^ n-1an = n & # 178 / 2 (1), then when n ≥ 2, there is a1 + 2A2 + 2 ^ 2A3 +... 2 ^ n-2an-1 = (n-1) & # 178 / 2 (2). Note that formula 2 has one less term than formula 1, then use formula 1 - 2 to get: 2 ^ n-1an = n & # 178 / 2 - (n-1) & # 178 / 2 = (2n-1) / 2, then an = (2n-1) /