# sequencesWatch

Thread starter 1 week ago
#1
In regards to question 2 on this paper, why is it 18000 times 0.8^3, surely it should be 18000 x 0.8^2 as n=3, as its the third year? https://pmt.physicsandmathstutor.com...c%20series.pdf
0
1 week ago
#2
(Original post by Bertybassett)
In regards to question 2 on this paper, why is it 18000 times 0.8^3, surely it should be 18000 x 0.8^2 as n=3, as its the third year? https://pmt.physicsandmathstutor.com...c%20series.pdf
After 1 year its 18,000*0.8
After 2 years its 18,000*0.8^2

If you're unsure, plug some initial values in to spot the pattern.
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Thread starter 1 week ago
#3
(Original post by mqb2766)
After 1 year its 18,000*0.8
After 2 years its 18,000*0.8^2

If you're unsure, plug some initial values in to spot the pattern.
that's why I dont understand why using the formula ar^(n-1) for the nth term (geometric) doesn't work for this.
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1 week ago
#4
(Original post by Bertybassett)
that's why I dont understand why using the formula ar^(n-1) for the nth term (geometric) doesn't work for this.
In that formula a is the value in year 1.
In this question 18,000 is the value in year 0. You should use a*r^n

Like I said, don't rely on just remembering formula, plug initial values in to make sure you understand.
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