Question

(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

Answer 1

Answer:
`\bold{\bigg(5,(1)/(81)\bigg)}`

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,

• 
  `a_4=(1)/(9)\cdot a_3\quad \rightarrow \quad a_4=(1)/(9)\cdot 1\quad \rightarrow \quad a_4=(1)/(9)\quad \rightarrow \quad \bigg(4, (1)/(9)\bigg)`
• 
  `a_5=(1)/(9)\cdot a_4\quad \rightarrow \quad a_5=(1)/(9)\cdot (1)/(9)\quad \rightarrow \quad a_5=(1)/(81)\quad \rightarrow \quad \bigg(5, (1)/(81)\bigg)`

Answer 2

a5 = 1/81

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