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sequences

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/download/Maths/A-level/C2/Topic-Qs/Edexcel-Set-1/C2%20%20Sequences%20and%20series%20-%20Geometric%20series.pdf
Reply 1
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/download/Maths/A-level/C2/Topic-Qs/Edexcel-Set-1/C2%20%20Sequences%20and%20series%20-%20Geometric%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.
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.
Reply 3
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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