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Math Help - Interest

  1. #1
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    Interest

    A fund of $2000 is to be accumulated by n annual payments of $150, plus a smaller final payment made one year after the nth payment. The interest rate is 4%. Find n and the amount of the final irregular payment.

    How do i find out what the smaller payment is? COuld someone show me how to work this out.
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  2. #2
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    Accumulation

    I have assumed that the first payment is immediate and the accumulation date is one year after the final smaller payment. You will need to modify my answer if this is not the case.

    Accumulated value of regular payments one year after the time of the final payment:
    150 * (1.04^(n+1) -1.04)/0.04 * 1.04
    Accumulated value of final payment one year after the time of the final payment:
    X * 1.04

    Simulataneous eqn

    150 * (1.04^(n+1) -1.04)/0.04 * (1.04) + X * 1.04 = 2000
    X < 150

    solving for n

    2000/1.04 - 150 * (1.04^(n+1) -1.04)/0.04 < 150
    1.51 < 1.04^(n+1)
    10.5<n+1
    n>9.5
    n=10
    X = 50.12
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  3. #3
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    Quote Originally Posted by winterjm View Post
    A fund of $2000 is to be accumulated by n annual payments of $150, plus a smaller final payment made one year after the nth payment. The interest rate is 4%. Find n and the amount of the final irregular payment.

    How do i find out what the smaller payment is? COuld someone show me how to work this out.
    You simply must learn how to deal with "Fundamental Principles". ALL problems can be broken down to these basics. That's why they are called "fundamental".

    First, get in the habit of writing clear and complete defintions. Let the notation help you.

    i = 0.04
    v = 1/(1+i) = 1/1.04
    w = 1/v = 1.04
    p = 150 -- The regular payment
    q = The last irregular payment

    Then the sum can be written (Again, I'll assume the first payment is immediate.)

    pw^{n}\;+\;pw^{n-1}\;+...+\;pw^{1}\;+\;q\;=\;2000

    Your ONLY task is to find that sum. There are various ways to do that. In this case, it is important to notice that it is not particularly possible with that silly 'q' in there. Solve this:

    pw^{n}\;+\;pw^{n-1}\;+...+\;pw^{1}\;=\;2000

    And figure out what to do with the fractional part of 'n'.

    Can you find a more convenient expression for the sum on the left-hand side? You need to be able to do that.
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  4. #4
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    Simplification

    The simplification in your notation is:

    p/i * [w^(n+1) - w]

    Although please note that in my earlier solution I assume that the final payment accumulates for a year.
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  5. #5
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    Quote Originally Posted by winterjm View Post
    A fund of $2000 is to be accumulated by n annual payments of $150, plus a smaller final payment made one year after the nth payment. The interest rate is 4%. Find n and the amount of the final irregular payment.

    How do i find out what the smaller payment is? COuld someone show me how to work this out.
    I would say there is a formula for this you have to work out, I would help you but what type of business math is this? College level?
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  6. #6
    GAMMA Mathematics
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    Quote Originally Posted by winterjm View Post
    A fund of $2000 is to be accumulated by n annual payments of $150, plus a smaller final payment made one year after the nth payment. The interest rate is 4%. Find n and the amount of the final irregular payment.

    How do i find out what the smaller payment is? COuld someone show me how to work this out.
    150s_{\overline{n|}}(1+i)+K=2000

    Two unknowns, but one equation. What to do?

    Solve for n without payment K:

    150s_{\overline{n|}}(1+i)=2000

    150\frac{(1+i)^n-1}{i}(1+i)=2000

    150\frac{(1.04)^n-1}{.04}(1.04)=2000

    1.04^n=1.5128

    .03922n=.413962

    n = 10.55

    n is the term of the repayments. It must be an integer, so therefore n=10

    Now, solve for the annuity at n=10.

    s_{\overline{10|}}=\frac{1.04^{10}-1}{.04}\Rightarrow 12.006

    150s_{\overline{10|}}(1.04) = 1872.95

    The balance of that and the 2000 lump sum is: K=127.05
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