Write the first five terms of the sequence defined by the explicit formula an=150(1/5)^n-1
150,30,6,6/5,6/25
Sure, here are the first five terms of the sequence:
1. a₁ = 150(1/5)^(1-1) = 150(1/5)^0 = 150(1) = 150
2. a₂ = 150(1/5)^(2-1) = 150(1/5) = 30
3. a₃ = 150(1/5)^(3-1) = 150(1/25) = 6
4. a₄ = 150(1/5)^(4-1) = 150(1/125) = 1.2
5. a₅ = 150(1/5)^(5-1) = 150(1/625) = 0.24
Remember, if this sequence was a person, it would be getting smaller and smaller just like my chances of becoming a professional unicycle juggler!
To find the first five terms of the sequence defined by the explicit formula an = 150(1/5)^(n-1), we can plug in the values of n from 1 to 5 and calculate the corresponding terms:
When n = 1:
a1 = 150(1/5)^(1-1)
= 150(1/5)^0
= 150 * 1
= 150
When n = 2:
a2 = 150(1/5)^(2-1)
= 150(1/5)^1
= 150 * (1/5)
= 150/5
= 30
When n = 3:
a3 = 150(1/5)^(3-1)
= 150(1/5)^2
= 150 * (1/5)^2
= 150 * (1/25)
= 150/25
= 6
When n = 4:
a4 = 150(1/5)^(4-1)
= 150(1/5)^3
= 150 * (1/5)^3
= 150 * (1/125)
= 150/125
= 1.2
When n = 5:
a5 = 150(1/5)^(5-1)
= 150(1/5)^4
= 150 * (1/5)^4
= 150 * (1/625)
= 150/625
= 0.24
So, the first five terms of the given sequence are: 150, 30, 6, 1.2, 0.24.
To find the first five terms of the sequence defined by the explicit formula an = 150(1/5)^(n-1), we need to substitute the values of n into the formula.
Let's calculate the first five terms one by one:
For n = 1: a1 = 150(1/5)^(1-1) = 150(1/5)^0 = 150 * 1 = 150
For n = 2: a2 = 150(1/5)^(2-1) = 150(1/5)^1 = 150 * 1/5 = 30
For n = 3: a3 = 150(1/5)^(3-1) = 150(1/5)^2 = 150 * 1/25 = 6
For n = 4: a4 = 150(1/5)^(4-1) = 150(1/5)^3 = 150 * 1/125 = 1.2
For n = 5: a5 = 150(1/5)^(5-1) = 150(1/5)^4 = 150 * 1/625 = 0.24
Thus, the first five terms of the sequence are: 150, 30, 6, 1.2, 0.24.
huh? This is algebra II. Just plug in 1..5 for n
a1 = 150(1/5)^(1-1) = 150(1/5)^0 = 150
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