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

1 Answer
Tony B
Nov 16, 2015

#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))#

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

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#

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