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Math Help - Investment Problem

  1. #1
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    Investment Problem

    Hey can someone plz help on this problem.


    Jason is 20 years old and his salary next year will be $20000. His salary is going to increase at a rate of 5% each year till he is 50. How much money would he have saved by that time if he saves 5% each year and invest the savings at an interest rate of 8%?


    Thanks

    JT
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  2. #2
    Junior Member
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    There are some possible interpretive variations in your fact-set that might need clarifying, such as number and timing of the investment-deposits. But you can easily adjust the following basic approach as need be.

    I'll set it up assuming Jason will make one deposit per year, consisting of 5% of his salary of that given year, and it will be made at the end of such year. Further, he'll have a total of 30 such deposits, and we'll compute his accumulated amount immediately following the final one.

    The set-up lays out as....

     T\ =\ 20,000(0.05)(1.08)^{29}\ +\ 20,000(1.05)(0.05)(1.08)^{28}\ +\ ...\ +\ 20,000(1.05)^{29}(0.05)

    ...where T is the total accumulated amount. Note that the FV of the first, second, and final deposit are shown explicitly, with the others implied.

    Note that this is a Geometric Series, with initial amount a =  20,000(0.05)(1.08)^{29} ; common ratio r =  (1.05)(1.08)^{-1} ; and there are a total of n = 30 terms.

    With that, use the sum-of-a-GS formula...

     T\ =\ \frac{a(1-r^n)}{1-r}

    ...to quickly obtain Jason's final accumulated total.
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  3. #3
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    Thanks alot your solution is really helpful.
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  4. #4
    MHF Contributor
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    Here's a way to "general case" your problem:

    A = accumulated total (?)
    S = initial salary (20000)
    i = interest rate paid (.08)
    j = rate (percentage) of salary deposited (.05)
    k = rate of salary increase (.05)
    n = number of years (30)

    A = Sk{[(1 + i)^n - (1 + j)^n] / [i - j]

    A = 20000(.05)[(1.08^30 - 1.05^30) / (.08 - .05)] = 191,357.15046....
    (same results as with LochWulf's way)
    Last edited by Wilmer; September 29th 2009 at 05:34 PM. Reason: none
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