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

##### 1 Answer
Nov 16, 2015

$r = \frac{1}{8}$
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 ${\left(- 1\right)}^{f \left(i\right)}$

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Consider the alternating +ve and -ve

If ${a}_{i} = {a}_{1}$ then we have${\left(- 1\right)}^{i + 1} k {r}^{i} = {\left(- 1\right)}^{2} k {r}^{1} = 128$
if${a}_{i} = {a}_{2}$ then we have ${\left(- 1\right)}^{i + 1} k {r}^{i} = {\left(- 1\right)}^{3} k {r}^{2} = \left(- 16\right)$

To find r ignore the ${\left(- 1\right)}^{i + 1}$ part remembering that this alternates the +ve and -ve. Thus the solution for r will be +ve
$\frac{k {r}^{2}}{k {r}^{1}} = r = \frac{16}{128} = \frac{1}{8}$

You are not asked to find the value of k.

But you could by

${\left(- 1\right)}^{2} k {\left(\frac{1}{8}\right)}^{1} = 128$

so $k = 128 \times \frac{8}{1} \div i \mathrm{de} 1 = 1024$

for a quick check I select ${a}_{3}$

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