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Math Help - Annuity reinvestment problem

  1. #1
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    Annuity reinvestment problem

    A man invests 1000 at the beginning of each year into a fund that pays an annual interest rate of 5.6%. The annual interest payments are deposited into a fund that pays 6.2% annually. What is his total accumulation at the end of 10 years?

    I thought that the way to solve this problem was:

    total accumulation = 56\"{s}_{10,0.062}+10,000=>56(\frac{1.06^{10}-1}{0.062})+10,000=10791.29

    But the answer in the back of my book says that the the annuity is increasing, but I don't understand how.

    The amount of interest for year k is 1000k(0.056)=56k. The accumulation of these payments plus interest is 56(Is)10,0.06=3,730.48. Total accumulation is 13,730.48.

    Can someone explain how that answer was achieved? Thanks.
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  2. #2
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    Re: Annuity reinvestment problem

    The 6.2% account:
    Code:
    YR   DEPOSIT   INTEREST  BALANCE
    1      56.00        .00    56.00
    2     112.00       3.47   168.47
    ...
    10    560.00     185.09  3730.4829.....
    Good nuff?
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  3. #3
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    Re: Annuity reinvestment problem

    Quote Originally Posted by downthesun01 View Post
    The amount of interest for year k is 1000k(0.056)=56k. The accumulation of these payments plus interest is 56(Is)10,0.06=3,730.48. Total accumulation is 13,730.48.
    u = 1st payment (56)
    v = constant payment increase (56)
    n = number of periods (10)
    i = interest rate per period (.062)
    r = 1 + i (1.062)
    f = future value (?)

    f = { i [u(r^n - 1) - v(n - 1)] + vr[r^(n - 1) - 1]} / i^2
    f = 3730.4829.....

    Above also handles problems like:
    Jack opens annuity account in which he deposits $500 as 1st annual payment,
    then increases payments by $50 each year....u = 500, v = 50
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  4. #4
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    Re: Annuity reinvestment problem

    Nevermind. I wasn't thinking. I the first account has 1000 added to it every year.
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