Question

Which sequence is modeled by the graph below?

Answer 1

Answer:

$a_n=27\left(\dfrac{1}{3}\right)^{n-1}$

Step-by-step explanation:

From inspection of the graph, the given points are:

• (2, 9)
• (3, 3)
• (4, 1)

If we draw a line through the given points, the line is a curve rather than a straight line. If the line was a straight line, the graph would be modeled as an arithmetic sequence. Therefore, as the line is a curve, the given points are modeling a geometric sequence.

General form of a geometric sequence:

 $a_n=ar^{n-1}$

where:

• a is the first term
• r is the common ratio
• $a_n$ is the nth term

Rewrite the given points as terms of the sequence:

• (2, 9)  ⇒  a₂ = 9
• (3, 3)  ⇒  a₃ = 3
• (4, 1)  ⇒  a₄ = 1

To find the common ratio r, divide consecutive terms:

$\implies r=\dfrac{a_3}{a_2}=\dfrac{3}{9}=\dfrac{1}{3}$

Calculate the first term (a) by substituting the found value of r and the given values of one of the terms into the formula:

$\implies a_2=9$

$\implies a\left(\dfrac{1}{3}\right)^{2-1}=9$

$\implies \dfrac{1}{3}a=9$

$\implies a=27$

Substitute the found values of r and a into the general formula to create the sequence modeled by the graph:

$a_n=27\left(\dfrac{1}{3}\right)^{n-1}$

Learn more about geometric sequences here:

brainly.com/question/25398220

brainly.com/question/27783194

Answer 2

• (2,9)
• (3,3)
• (4,1)

We can observe

• 9/3=3
• 3/3=1

Its a geometric progression having first term 9(3)=27 and common ratio as 1/3

So

The firmula is

• a_n=27(1/3)^{n-1}

Option C