Question

What is the common ratio of the geometric sequence: 128, -16, 2, -1/4, …?

Answer 1

`r = 1/8`
Explanation also solves the rest of the sequence structure!

Explanation:

where `a_1=128 ;color(white)(xx) a_2 = (-16) ; color(white)(xx)a_3=2 color(white)(xx)and a_4=(-1/4)...`

Let any term in this sequence be `a_i`
Let a constant be k
let the ratio be r

Two points to note:

Point 1:
it is reducing so `r<1`

Point 2:
The sequence alternates positive to negative so involves `(-1)^(f(i))`

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Consider the alternating +ve and -ve

If `a_i = a_1` then we have`(-1)^(i+1) kr^i = (-1)^2 kr^1 =128`
if`a_i =a_2` then we have `(-1)^(i+1)kr^i = (-1)^3 kr^2 = (-16)`

To find r ignore the `(-1)^(i+1)` part remembering that this alternates the +ve and -ve. Thus the solution for r will be +ve
`(kr^2)/(kr^1) = r = 16/128 = 1/8`

You are not asked to find the value of k.

But you could by

`(-1)^(2)k( 1/8 )^1= 128`

so `k =128 times 8/1 divide 1 = 1024`

for a quick check I select `a_3`

`a_3 = 2 -> ? [ (-1)^(3+1) (1024) times (1/8)^3 ] = 2`