The common ratio of the geometric sequence is 1/3
How to determine the common ratio?
The terms are given as:
T2 = 12
T4 = 4/3
The nth term of a geometric sequence is calculated using
Tn = ar^(n-1)
So, we have:
T2 = ar
T4 = ar^3
Divide T4 by T2
[tex]\frac{T4}{T2} = \frac{ar^3}{ar}[/tex]
Divide
[tex]\frac{T4}{T2} = r^2[/tex]
Substitute T2 = 12 and T4 = 4/3
[tex]\frac{4/3}{12} = r^2[/tex]
Divide
[tex]\frac{4}{36} = r^2[/tex]
Take the square roots
[tex]r = \frac 26[/tex]
Simplify
[tex]r = \frac 13[/tex]
Hence, the common ratio is 1/3
Read more about geometric sequence at:
https://brainly.com/question/24643676
#SPJ5
