If A3 = - 4, A6 = 4, then A9=

If A3 = - 4, A6 = 4, then A9=

a3=a1q^2=-4 (1)
a6=a1q^5=4 (2)
(2) (1) get
q^3=-1
q=-1
a1=-4
a9=-4*(-1)^8=-4

In the proportional sequence an, A3 and A8 are the equations 3x ^ 2_ Kx-7 = 0, and (A3 + A8) ^ 2 = - 3a4a7 + 2, then the value of K is

There are A3 + A8 = - B / A, A3 * A8 = C / A and A4 * A7 = A3 * A8 in the equal ratio sequence
It's easy to solve K by substituting all of them

If A3 = - 9, a7 = - 1, then A5 = ()
A 3 or-3 B3 C-3 d above are all wrong, answer and explain the reason. Thank you

C
a3=—9=a1q^2
a7=—1=a1q^6
Divide 2 by 1
So Q ^ 4 = 1 / 9
So Q ^ 2 = 1 / 3
From Formula 1, A1 = - 27
So A5 = A1 × Q ^ 4 = - 27 × 1 / 9 = - 3

If A3 = - 9, a7 = - 1, then the value of A5 is ()
A. 3 or - 3B. 3C. - 3D. - 5

In the sequence {an}, A3 = - 9, a7 = - 1. According to the definition and properties of the sequence, we can get a52 = A3 · A7 = 9, and get A5 = - 3, or A5 = 3. If A5 = 3, then A42 = A3 · A5 = - 27, A4 does not exist