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Math Help - quick interest question ...

  1. #1
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    quick interest question ...

    To provide an annual scholarship for 25 years, a donation of $50,000 is invested in an account for a scholarship that will start a year after the investment is made. If the money is invested at 5.5% per annum, compounded annually; determine the amount of each scholarship?

    I would greatly appreciate it if I could be given some help.
    Thanks so much in advance.
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  2. #2
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    Quote Originally Posted by asiankatt View Post
    To provide an annual scholarship for 25 years, a donation of $50,000 is invested in an account for a scholarship that will start a year after the investment is made. If the money is invested at 5.5% per annum, compounded annually; determine the amount of each scholarship?

    I would greatly appreciate it if I could be given some help.
    Thanks so much in advance.
    I don't know if there is a ready formula for this. Let us derive one.

    Let X = amount of scholarship to be deducted every year
    A = amount of investment after any year
    P = principal = $50,000 here.
    r = rate of interest = 0.055 here.
    n = number of years

    In compounded annually,
    A = P(1+r)^n

    After year 1,
    A = P(1+r)
    Atfer withdrawal of X,
    A = P(1+r) -X

    After year 2,
    A = [P(1+r) -X](1+r) = P(1+r)^2 -X(1+r)
    After withdrawal of X,
    A = P(1+r)^2 -X(1+r) -X

    After year 3,
    A = [P(1+r)^2 -X(1+r) -X](1+r)
    A = P(1+r)^3 -X[(1+r)^2 +(1+r)]
    After withdrawal of X,
    A = P(1+r)^3 -X[(1+r)^2 +(1+r)] -X
    .
    .
    After year 25,
    A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)]
    After withdrawal of X,
    A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] -X ------(i)
    And that is now equal to zero.

    The [(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] can be rewritten as
    [(1+r) +(1+r)^2 +(1+r)^3 +...+(1+r)^24].
    It is a geometric series where
    common ratio = (1+r)
    a1 = (1+r) also
    n = 24
    So, since (1+r) = (1+0.055) = 1.055, then,
    Sn = (a1)[(1 -r^n)/(1-r)]
    S(24) = (1.055)[(1 -(1.055)^24) / (1 -(1.055)]
    S(24) = (1.055)[(-2.61459)/(-0.055)]
    S(24) = 50.15259 ----------------------***

    Substituting that and the givens into (i),
    A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] -X ------(i)
    which is zero, so,
    0 = (50,000)(1.055)^25 -X(50.15259) -X
    0 = 190,670 -X(51.15259)
    X = (190,670)/(51.15259)
    X = $3727.47 -----------------answer.
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  3. #3
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    Hello, ticbol!


    There is a "Sinking Fund" formula . . . which you derived from scratch.

    . . Lovely work!


    One of its forms looks like this: . A \;=\;P\frac{i(1 + i)^n}{(1+i)^n - 1}

    where: . \begin{array}{cccc} P & = & \text{principal invested} \\ i & = & \text{periodic interest rate} \\ n & = & \text{number of periods} \\ A & = & \text{periodic withdrawl} \end{array}

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, ticbol!


    There is a "Sinking Fund" formula . . . which you derived from scratch.

    . . Lovely work!


    One of its forms looks like this: . A \;=\;P\frac{i(1 + i)^n}{(1+i)^n - 1}

    where: . \begin{array}{cccc} P & = & \text{principal invested} \\ i & = & \text{periodic interest rate} \\ n & = & \text{number of periods} \\ A & = & \text{periodic withdrawl} \end{array}

    Good morning, Soroban.

    So that is a Sinking Fund. Umm.
    I just had time last night to play with it. I thought the question was interesting.
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  5. #5
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    Hello again, ticbol!

    I can appreciate the reasoning and the algebra
    . . that went into your derivation.

    Every year or so, I can't my list of favorite formulas,
    . . and I've been forced to derive the Amortization Formula.
    So I've had considerable practice.

    If that was your first attempt at such a derivation, well done!
    . . And an excellent explanation, too.

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