If the sequence {an} is an equal ratio sequence composed of positive numbers and the common ratio is not 1, then the size relationship between a1 + A8 and A4 + A5 is () A. A1 + A8 ＞ A4 + a5b. A1 + A8 ＜ A4 + a5c. A1 + A8 = A4 + a5d

If the sequence {an} is an equal ratio sequence composed of positive numbers and the common ratio is not 1, then the size relationship between a1 + A8 and A4 + A5 is () A. A1 + A8 ＞ A4 + a5b. A1 + A8 ＜ A4 + a5c. A1 + A8 = A4 + a5d

∵ equal ratio sequence {an}, each item is a positive number ∵ A1 > 0, Q > 0 and Q ≠ 1A1 + A8 - (A4 + A5) = (a1 + a1q7) - (a1q3 + a1q4) = A1 (q3-1) (q4-1) > 0 & nbsp;; a1 + A8 > A4 + A5, so select a

In the arithmetic sequence {an}, D ≠ 0, and A1, A2, A4 are equal ratio sequence, find (a1 + A2 + A4) / (A2 + A4 + A8)

A1, A2, A4 are equal ratio sequence
(a2)^2=a1*a4
(a2)^2=(a2-d)(a2+2d)
(a2)^2=(a2)^2+a2d-2d^2
a2d=2d^2
a2=2d
(a1+a2+a4)／(a2+a4+a8)
=(a2-d+a2+a2+2d)/(a2+a2+2d+a2+6d)
=(3a2+d)/(3a2+8d)
=(3*2d+d)/(3*2d+8d)
=7d/(14d)
=1/2

In the sequence {an}, A1, A2, A3 are equal difference sequence, and their sum is 15, A4, A5, A6 are equal ratio sequence, and their product is 27
For any natural number n, a (n + 6) = an
(1) Find A2, A5, a8
(2) Find the first 101 terms and S101 of the sequence {a (3K-1)} (k is a natural number)

If a 1, a 2 and a 3 are equal difference sequence, then 2 * a 2 = a 1 + a 3, then 3 a 2 = 15 and a 2 = 5
A4, A5 and A6 are in equal proportion sequence, that is, the square of A5 = A4 * A6
Then the cube of A5 = 27, then A5 = 3
Because a (n + 6) = an, then a (2 + 6) = A2, that is, a8 = A2 = 5
(2) The sum of the first 101 terms of {a (3K-1)} (k is a natural number) is the sum of A2, A5, A8, a11.a302
These numbers can be divided into two types. One is A2, A8, A14, A20. There are 51 numbers in A302. From a (n + 6) = an, we know that each number is equal to A2 and is 5
The other is A5, a11, A17. A299. There are 50 numbers in total, and each number is equal to A5 and 3
s101=51*5+50*3=405

The {an} is an equal ratio sequence, an ＞ 0, Q ≠ 1, A1 * A8 = A4 * A5, ask the relationship between a1 + A8 and A4 + A5

A1 + A8 = A1 (1 + Q ^ 7) A4 + A5 = A1 (Q ^ 3 + Q ^ 4) compare the size of (1 + Q ^ 7) and (Q ^ 3 + Q ^ 4), that is, (Q ^ 3-1) (Q ^ 4-1) = (Q ^ 7 + 1) - (Q ^ 3 + Q ^ 4) no matter Q ＞ 1 or Q ＜ 1 (Q ^ 3-1) (Q ^ 4-1) ＞ 0, then (Q ^ 7 + 1) - (Q ^ 3 + Q ^ 4) ＞ 0, that is, (Q ^ 7 + 1) ＞ (Q ^ 3 + Q ^ 4), then a1 + A8 ＞ A4 + A5