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(98 Points!!) Can someone take a look at this? I'm stumped. (Answer choices provided)

What is the fifth term of the geometric sequence?

Geometric sequence represented by a graph in quadrant 1 with n on x axis and a sub n on y axis. Points at (1,81), (2,9), and (3,1). (Graph attached)

Hint: an = a1(r)n − 1, where a1 is the first term and r is the common ratio

98 Points Can someone take a look at this Im stumped Answer choices provided What is the fifth term of the geometric sequence Geometric sequence represented by class=

Respuesta :

wegnerkolmp2741o

Answer:

a5 = 1/81

Step-by-step explanation:

(1,81) (2,9) (3,1)

The sequence goes

81,9,1,....

We are multiplying by 1/9 each time

a1 = 81

r= 1/9

an = a1 r^(n-1)

an = 81 * (1/9) ^ (n-1)

The 5th terms

a5 = 81 * (1/9) ^ (5-1)

a5 = 81 * (1/9) ^4

a5 = 81/6561

a5 = 1/81

tramserran

Answer:   [tex]\bold{\bigg(5,\dfrac{1}{81}\bigg)}[/tex]

Step-by-step explanation:

(1, 81), (2, 9), (3, 1) means

  • a₁ = 81
  • a₂ = 9
  • a₃ = 1

the common ratio (r) is: 9 ÷ 81  →  r =  1/9

therefore,

  • [tex]a_4=\dfrac{1}{9}\cdot a_3\quad \rightarrow \quad a_4=\dfrac{1}{9}\cdot 1\quad \rightarrow \quad a_4=\dfrac{1}{9}\quad \rightarrow \quad \bigg(4, \dfrac{1}{9}\bigg)[/tex]
  • [tex]a_5=\dfrac{1}{9}\cdot a_4\quad \rightarrow \quad a_5=\dfrac{1}{9}\cdot \dfrac{1}{9}\quad \rightarrow \quad a_5=\dfrac{1}{81}\quad \rightarrow \quad \bigg(5, \dfrac{1}{81}\bigg)[/tex]

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