If A5 = 4 in {an}, then A2 · A8 is equal to

If A5 = 4 in {an}, then A2 · A8 is equal to

a2=a1×q
a8=a1×q^7
And A5 = A1 × Q ^ 4
So A2 × A8 = (A1 ^ 2) × (Q ^ 8) = A5 ^ 2 = 4 ^ 2 = 16
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Given the sequence of equal ratio {an}, a1 + a3 = 5, A3 + A5 = 20, then the general formula of {an} is______ ．

Let the common ratio of the equal ratio sequence {an} be q, then 20 = A3 + A5 = Q2 (a1 + a3) = 5q2, ∧ Q2 = 4, ∧ q = ± 2, substituting into a1 + a3 = 5, we get A1 = 1, when q = 2, an = 2N-1; when q = - 2, an = (- 2) n-1. So the answer is an = 2N-1 or an = (- 2) n-1

It is known that A1 = 27, A9 = 1 / 243, Q

a9=a1*q^8
So Q ^ 8 = (1 / 243) / 27 = 1 / 3 ^ 8
q

In the equal ratio sequence {an}, A1 = half, A9 = 128,

a9=a1*q^8 q^8=256 q=2
S8=a1(1-q^8)/(1-q)=127.5