Answer:
[tex]a_{1}[/tex] = 1/2
[tex]a_{n}[/tex] = [tex]a_{n-1}[/tex] * [tex]\frac{4}{3}[/tex]
Step-by-step explanation:
plug 1 into [tex]a_{n}[/tex]=1/2(4/3)^n−1 to find [tex]a_{1}[/tex]
=1/2(4/3)^1−1
[tex]a_{1}[/tex] = 1/2 (4/3)^0
[tex]a_{1}[/tex] = 1/2 * 1
[tex]a_{1}[/tex] = 1/2
[tex]a_{n}[/tex] = [tex]a_{n-1}[/tex] * r where r is the ratio
since (4/3) is being taken to an exponent, (4/3) is r
[tex]a_{n}[/tex] = [tex]a_{n-1}[/tex] * [tex]\frac{4}{3}[/tex]
The recursive rule for the geometric sequence is given by [tex]\rm a_n = a_{n-1}\times \dfrac{4}{3}[/tex] and this can be determined by finding the geometric ratio 'r'.
Given :
Geometric Sequence -- [tex]\rm a_n = \dfrac{1}{2}\left(\dfrac{4}{3}\right)^{n-1}[/tex]
Now, substitute the value of n = 1 in the above given geometric sequence.
[tex]\rm a_1 = \dfrac{1}{2}\left(\dfrac{4}{3}\right)^{1-1}[/tex]
[tex]\rm a_1 = \dfrac{1}{2}\left(\dfrac{4}{3}\right)^{0}[/tex]
[tex]\rm a_1 = \dfrac{1}{2}[/tex]
Now, substitute the value of n = 2 in the given geometric sequence.
[tex]\rm a_2 = \dfrac{1}{2}\left(\dfrac{4}{3}\right)^{2-1}[/tex]
[tex]\rm a_2 = \dfrac{1}{2}\left(\dfrac{4}{3}\right)^{1}[/tex]
[tex]\rm a_2 = \dfrac{2}{3}[/tex]
To determine the value of r which is the geometric ratio use the below calculation:
[tex]\rm r = \dfrac{a_2}{a_1}=\dfrac{\dfrac{2}{3}}{\dfrac{1}{2}}[/tex]
[tex]\rm r = \dfrac{4}{3}[/tex]
So, the recursive rule for the geometric sequence is given by:
[tex]\rm a_n = a_{n-1}\times \dfrac{4}{3}[/tex]
For more information, refer to the link given below:
https://brainly.com/question/24357959
