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Sagot :
Thisbis a general geometric progression
- first term=a=9
- common ratio=2/3=r
So iterative formula
- ar^{n-1}
- 9(2/3)^{n-1}
Answer:
[tex]a_n=9 \cdot \left(\frac{2}{3}\right)^{n-1}[/tex]
Step-by-step explanation:
Given:
[tex]\begin{cases}a_1=9\\a_n=\frac{2}{3}(a_n-1)\end{cases}[/tex]
General form of a geometric sequence:
[tex]a_n=ar^{n-1}[/tex]
where:
- a is the first term
- r is the common ratio
Therefore:
- [tex]a = a_1 = 9[/tex]
- [tex]r = \dfrac{2}{3}[/tex]
Substitute the given values into the general formula:
[tex]\implies a_n=9 \cdot \left(\frac{2}{3}\right)^{n-1}[/tex]
Learn more at:
https://brainly.com/question/27783194
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