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Math Help - Monthly Payments + Interest Problem

  1. #1
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    Monthly Payments + Interest Problem

    Hello!

    I have a problem that I'm trying to figure out how to do the steps of. I have the solution, but I can't figure out how to get to it. The problem is as follows:

    "Gina wants to establish a college fund for her newborn daughter that will have accumulated $120,000 at the end of 18 yr. If she can count on an interest rate of 6%, compounded monthly, how much should she deposit each month to accomplish this?"

    The answer is $309.79, but I don't see how. Can someone please explain the steps to me? Thanks in advance.
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  2. #2
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    There's a formula for that sort of thing, but feel free to build it from basic principles.

    n = 18 years = 18*12 months = 216 months

    i = 0.06

    i^{(12)} = 0.06/12 = 0.005

    Monthly Accumulation Factor: a = 1+i^{(12)} = 1.005

    Monthly Deposit: P <== This is what we seek.

    Final Accumulation: A = $120,000

    If you had to do it in one payment, what would it look like?

    P*a = 120000 and P = 120000/1.005 = 119402.99

    If you had to do it in two payments, what would it look like?

    P*a^{2} + P*a = 120000 = P(a^{2} + a) = P(1.010025 + 1.005) = P(2.015025) and P = 120000/2.015025 = 59.552.62

    It is hoped that you can see into the past and answer this last question, what would it look like if you have 216 months?

    P(a^{216} + a^{215} + a^{214} + ... + a^{2} + a) = 120000

    Your task is to add up all that stuff in the parentheses. It is a finite geometric series. Can you finish?

    I get $308.26, so perhaps we have not defined the problem correctly.
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  3. #3
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    The formula for these types of problems is as follows

    For the balance at the time of the last deposit:

    Future Value = (C/r)[((1+r)^n) - 1]

    Where C is your deposit, n is the number of deposits and r is the relevant interest rate.

    C = ?
    r = 0.06/12 = 0.005
    n = 12 x 18 = 216
    FV = 120000

    120000 = (C/0.005)[(1.005^216) - 1]

    Solve for C and you get $309.79.

    As for not using the formula, you can think of the 120000 as the future value of all the deposits. I.e., the first deposit will have a value of C(1.005)^216, 2nd will have a value of C(1.005)^215, etc... Add them all up and it will total 120000.

    It can be expressed as a sum of a geometric series. Use the formula

    S = (a(r^n - 1))/(r - 1)

    with S = 120000
    r = 1.005
    n = 216

    Solve for A and you get 309.79
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  4. #4
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    Thanks so much, Tiger_shark! I spent all afternoon trying to figure that out. The book I have didn't list anything remotely close to that formula, so it was killing me to figure out what it was. Thanks once again!
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