Two terms in a geometric sequence are a5=15 and a6=1. What is the recursive rule that describes the sequence?

A) a1=225; an=an−1⋅5 an=an−1⋅5 , ,

B) a1=11,390,625; an=an−1⋅15 an=an−1⋅15 , ,

C) a1=50,625; an=an−1⋅15 an=an−1⋅15 , ,

D) a1=759,375; an=an−1⋅115

Please help!!

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~

the image is given below

since the options are not given so we draw it from the equation

the given equation is

y = (x-1) (x+4)

y = y² + 3x - 4

the above equation is a quadratic equation which will have two roots. if we draw the graph for it, it will look like this

"standard form:

3x2 + 14x − 10 = 0

solutions based on quadratic formula:

x1   = −14 − √ 142 − 4×3×(−10) /   2×3   = −14 − 2 × √ 79 /   6   ≈ −5.296

x2   = −14 + √ 142 − 4×3×(−10) /   2×3   = −14 + 2 × √ 79 /   6   ≈ 0.629