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    Andy invests £A at a rate if interest 4% per annum. After five years it will be worth £10,000. What is A?

    My working out:
    Un = AR^n-1
    10,000 = A * 1.04^(5-1)
    A = 10000/1.04^(5-1)

    However, the book says the answer is:
    A = 10000/1.04^(5)

    I thought the formula was R to the the power of n-1, by the book only uses R to the power of n. Why is this?
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    (Original post by Desultory)
    Andy invests £A at a rate if interest 4% per annum. After five years it will be worth £10,000. What is A?

    My working out:
    Un = AR^n-1
    10,000 = A * 1.04^(5-1)
    A = 10000/1.04^(5-1)

    However, the book says the answer is:
    A = 10000/1.04^(5)

    I thought the formula was R to the the power of n-1, by the book only uses R to the power of n. Why is this?
    Depends how you contextualise your n.

    What is n in your context?

    Approach 1:

    After 1 year, the worth W is A\cdot 1.04

    After 2 years, it has worth W=A \cdot 1.04^2

    ...

    After n year, it has worth W=A \cdot 1.04^n

    Approach 2:

    If you want to use n-1, you would have to say something like "In Year 1, we have W=A, then in Year 2 we have W=A\cdot 1.04, then in Year n we have W=A \cdot 1.04^{n-1}" but even then, the phrase 'after 5 years' implies you're looking for the value right after Year 5 ends, so what's the value in Y6? Plug in n=6 and you're back to the first scenario.
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    (Original post by Desultory)
    Andy invests £A at a rate if interest 4% per annum. After five years it will be worth £10,000. What is A?

    My working out:
    Un = AR^n-1
    10,000 = A * 1.04^(5-1)
    A = 10000/1.04^(5-1)

    However, the book says the answer is:
    A = 10000/1.04^(5)

    I thought the formula was R to the the power of n-1, by the book only uses R to the power of n. Why is this?
    The question says after 5 years, meaning that it take 5 years for it to reach £10,000, this then means that the value for n would be 6 as you start at year 0 and 5 whole years later would be year 5, 0,1,2,3,4,5 is 6 so N-1 = 5
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    (Original post by RDKGames)
    Depends how you contextualise your n.

    What is n in your context?

    Approach 1:

    After 1 year, the worth W is A\cdot 1.04

    After 2 years, it has worth W=A \cdot 1.04^2

    ...

    After n year, it has worth W=A \cdot 1.04^n

    Approach 2:

    If you want to use n-1, you would have to say something like "In Year 1, we have W=A, then in Year 2 we have W=A\cdot 1.04, then in Year n we have W=A \cdot 1.04^{n-1}" but even then, the phrase 'after 5 years' implies you're looking for the value right after Year 5 ends, so what's the value in Y6? Plug in n=6 and you're back to the first scenario.
    So what you are saying is that the formula given in the book for c2 is wrong and shouldnt be used?
    Or how do i know when to use it and when not to
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    (Original post by Desultory)
    So what you are saying is that the formula given in the book for c2 is wrong and shouldnt be used?
    Or how do i know when to use it and when not to
    No that's not what I'm saying in the slightest. I'm saying it depends how you lay out your scenario. I gave you two clear approaches in how it could've been layed out by you and how the n and n-1 can be used.
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    (Original post by RDKGames)
    No that's not what I'm saying in the slightest. I'm saying it depends how you lay out your scenario. I gave you two clear approaches in how it could've been layed out by you and how the n and n-1 can be used.
    Okay so why is the n-1 in the formula? I'm not sure I understand as in some other examples such as what is the 10th term in sequence 3 6 12 24, if you dont do ar^n-1, then you get the answer wrong
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    (Original post by Desultory)
    Okay so why is the n-1 in the formula? I'm not sure I understand as in some other examples such as what is the 10th term in sequence 3 6 12 24, if you dont do ar^n-1, then you get the answer wrong
    Because, and I'll use sequences as an example, u_n=ar^{n-1} gives you the nth term of a sequence so it makes sense.

    u_1=a
    u_2=ar
    u_3=ar^2
    ...
    u_n=ar^{n-1}

    Using u_r=ar^n would start you off from ar then and not a, unless you start from the 0th term - this is where the difference is! It's about where you start off, so to speak.

    And the question above shows this.
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    (Original post by RDKGames)
    Because, and I'll use sequences as an example, u_n=ar^{n-1} gives you the nth term of a sequence so it makes sense.

    u_1=a
    u_2=ar
    u_3=ar^2
    ...
    u_n=ar^{n-1}

    Using u_r=ar^n would start you off from ar then and not a, unless you start from the 0th term - this is where the difference is! It's about where you start off, so to speak.

    And the question above shows this.
    Okay, I kind of understand, but why are we starting on the second term in this quqestion then?
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    (Original post by Desultory)
    Okay, I kind of understand, but why are we starting on the second term in this quqestion then?
    Because we are using the 'after nth year' terminology and not 'during nth year' terminology (though there is nothing from stopping you from using it)

    So saying 'After 1 year' would mean you are on your second year, so you'd use n=2 in your W=A\cdot 1.04^{n-1} formula - and this fits perfectly with the context, as you'd expect to have the interest added after 1 year.
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    (Original post by RDKGames)
    Because we are using the 'after nth year' terminology and not 'during nth year' terminology (though there is nothing from stopping you from using it)

    So saying 'After 1 year' would mean you are on your second year, so you'd use n=2 in your W=A\cdot 1.04^{n-1} formula - and this fits perfectly with the context, as you'd expect to have the interest added after 1 year.
    Ok thanks, I think i understand now
 
 
 
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