In {an} where all items are positive proportional sequence, A2 * A4 = 4, a1 + A2 + a3 = 14, then the value of the largest positive integer n with an + an + 1 + an + 2 > 1 / 9 is satisfied?

In {an} where all items are positive proportional sequence, A2 * A4 = 4, a1 + A2 + a3 = 14, then the value of the largest positive integer n with an + an + 1 + an + 2 > 1 / 9 is satisfied?

∵a2*a4=4
∴a3=2.
q=1/2.
an=2^(4-n)
2^(9-3n)>1/9.
9-3n>=-3
n

In {an} where all items are positive proportional sequence, A2 + A4 = 4, a1 + A2 + a3 = 14, then the value of the largest positive integer n with an + an + 1 + an + 2 > 1 / 9 is satisfied?

a1(q+q^3)=4
a1(1+q+q^2)=14
Division of two formulas: (Q + Q ^ 3) / (1 + Q + Q ^ 2) = 2 / 7
Find Q
an+an+1+an+2=（a1+a2+a3）*q^(n-1)>1/9
The key is to find Q
To be honest, I didn't work out Q. I don't know if it's a calculation problem or something. I can only give you this idea for reference·······

If A4 = 2 and A5 = 5 in an, then the sum of the first 8 terms of (lgan) is equal to

In the equal ratio sequence, A1 * A8 = A2 * A7 = A3 * A6 = A4 * A5
lga1+lga2+… +lga8=lg(a1*a2*a3*… *a8)
=lg[(a1*a8)*(a2*a7)*… *(a4*a5)]
=lg(a4*a5)^4
=4lg(a4*a5)
=4×lg(2×5)
=4

Let the positive number sequence {an} be an equal ratio sequence, and A2 = 4, A4 = 16 +lga2nn．

Let the common ratio of sequence {an} be q, obviously Q ≠ 1, a4a2 = Q2 = 4, because an ＞ 0, n ∈ n, ∧ q = 2, A1 = a2q = 2, ∧ an = a1qn-1 = 2n, so lgan + 1 + lgan + 2 + +lga2nn=lg2n+1+lg2n+2+… +lg22nn2=[(n+1)+(n+2)+… +2n] n2lg2 = 3N2 + n2n2 · LG2, the original formula = LiMn →∞ (3N2 + n2n2 · LG2) & nbsp; = LG2 · LiMn →∞ 3N2 + n2n2 = 32lg2