Answer :
PART A
The geometric sequence is defined by the equation
[tex]a_{n}=3^{3-n}[/tex]
To find the first three terms, we put n=1,2,3
When n=1,
[tex]a_{1}=3^{3-1}[/tex]
[tex]a_{1}=3^{2}[/tex]
[tex]a_{1}=9[/tex]
When n=2,
[tex]a_{2}=3^{3-2}[/tex]
[tex]a_{2}=3^{1}[/tex]
[tex]a_{2}=3[/tex]
When n=3
[tex]a_{3}=3^{3-3}[/tex]
[tex]a_{3}=3^{0}[/tex]
[tex]a_{1}=1[/tex]
The first three terms are,
[tex]9,3,1[/tex]
PART B
The common ratio can be found using any two consecutive terms.
The common ratio is given by,
[tex]
r= \frac{a_{2}}{a_{1}} [/tex]
[tex]r = \frac{3}{9} [/tex]
[tex]r = \frac{1}{3} [/tex]
PART C
To find
[tex]a_{11}[/tex]
We substitute n=11 into the equation of the geometric sequence.
[tex]a_{11} = {3}^{3 - 11} [/tex]
This implies that,
[tex]a_{11} = {3}^{ - 8} [/tex]
[tex]a_{11} = \frac{1}{ {3}^{8} } [/tex]
[tex]a_{11}=\frac{1}{6561}[/tex]
The geometric sequence is defined by the equation
[tex]a_{n}=3^{3-n}[/tex]
To find the first three terms, we put n=1,2,3
When n=1,
[tex]a_{1}=3^{3-1}[/tex]
[tex]a_{1}=3^{2}[/tex]
[tex]a_{1}=9[/tex]
When n=2,
[tex]a_{2}=3^{3-2}[/tex]
[tex]a_{2}=3^{1}[/tex]
[tex]a_{2}=3[/tex]
When n=3
[tex]a_{3}=3^{3-3}[/tex]
[tex]a_{3}=3^{0}[/tex]
[tex]a_{1}=1[/tex]
The first three terms are,
[tex]9,3,1[/tex]
PART B
The common ratio can be found using any two consecutive terms.
The common ratio is given by,
[tex]
r= \frac{a_{2}}{a_{1}} [/tex]
[tex]r = \frac{3}{9} [/tex]
[tex]r = \frac{1}{3} [/tex]
PART C
To find
[tex]a_{11}[/tex]
We substitute n=11 into the equation of the geometric sequence.
[tex]a_{11} = {3}^{3 - 11} [/tex]
This implies that,
[tex]a_{11} = {3}^{ - 8} [/tex]
[tex]a_{11} = \frac{1}{ {3}^{8} } [/tex]
[tex]a_{11}=\frac{1}{6561}[/tex]