Find A5 from the equal ratio sequence {an}, a1 + A4 = 27, S6 = 189

Find A5 from the equal ratio sequence {an}, a1 + A4 = 27, S6 = 189

a1+a4=27 s6=189
a1(1+q^3)=27
The solution is A1 = 3, q = 2
a5=3*2^4=48

In the equal ratio sequence {an}, A2 = 2, A5 = 128. (1) find the general term formula of the sequence {an}; (2) let BN = log4an, find the sum of the first n terms of the sequence {BN}

(1) Let the tolerance of the sequence be D, from A2 = 2, A5 = 128, we can get A5 = A2 · Q3 {Q3 = 64} q = 4, an = A2 · QN − 2 = 22n − 3. (6 points) (2) ∵ BN = log4an, ∵ BN = log422n − 3 = 2n − 32, (10 points) ∵ Sn = B1 + B2 + +BN = n [− 12 + 2n − 32] 2 = N2 − 2n2. (14 points)

In {an}, A5 = 4, a7 = 6, then A9=______ ．

In the sequence {an}, A5 = 4, a7 = 6, then from the definition and properties of the sequence, we can get Q2 = a7a5 = 64 = 32, ｜ A9 = A7 · Q2 = 6 × 32 = 9, so the answer is 9

In the known arithmetic sequence {an}, A2 = - 20, a1 + A9 = - 28, if the sequence {BN} satisfies an = log2bn, let TN = b1b2... BN, and TN = 1, find the value of n

a2=a1+d=-20
a1+a9=a1+a1+8d=2a1+8d=-28
Solve the equation and get the result
a1=-22,d=2
an=-22+(n-1)x2=2n-24
an=log2bn=2n-24
bn=2^(2n-24)
Tn=b1b2...bn=2^[2(1+2+3+...+n)-24n]
=2^[2n(n+1)/2-24n]
=2^(n^2-23n)
Because TN = 1, so n ^ 2-23n = 0, so n = 0 or 23
N cannot be zero, so n = 23