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

1 Answer
Tony B
Nov 16, 2015

r = 1/8r=18
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))

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Consider the alternating +ve and -ve

If a_i = a_1 then we have (-1)^(i+1) kr^i = (-1)^2 kr^1 =128
ifa_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

Related questions
See all questions in Geometric Sequences
Impact of this question
3749 views around the world
You can reuse this answer
Creative Commons License