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Thread: Word Problem Reated to Compound Interest & Geometric Sequence

  1. #1
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    Question Word Problem Reated to Compound Interest & Geometric Sequence

    The question that I am struggling with is as follows:

    On the day of her birth, 1st January 1998, Mary's grandparents invested $x in a savings account. They continued to deposit $x on the first day of each month thereafter. The account paid a fixed rate of 0.4% interest per month. The interest was calculated on the last day of each month and added to the account. Let $A_n be the amount in Mary's account on the last day of the nth month, immediately after the interest had been added.

    As soon as Mary was 18 she decided to invest $15000 of thismoney in an account of the same type earning 0.4% interest per month. She withdraws $1000 every year on her birthday to buy herself a present. Determine how long it will take until there is no money in the account.

    There are 5 questions (Question A to E) related to thequestion described in the first paragraph above. I was able to answer and understand all the questions apart from the last one, which is described in the second paragraph above.

    The mark scheme says:
    r = 1.004^12 (= 1.049...)
    15000r^n - [1000(r^n-1)]/(r-1) = 0
    n = 27.8...
    So n = 28

    My understanding of this working above is that "r"is the common ratio of the geometric sequence for 1 year, "15000r^n"gives us the amount of interest plus principal you can gain over "n" years, and "[1000(r^n-1)]/(r-1)" gives us the total amount that you withdraw over the same period.

    Correct?

    However, I do not understand why the total amount forwithdrawal is "[1000(r^n-1)]/(r-1)" because the question says "She withdraws $1000 every year on her birthday...".So I do not understand why it is not simply"1000n" because it is a fixed amount of $1000 annually. Obviously "[1000(r^n-1)]/(r-1)" is the sum of geometric sequence with the first term of 1000 and common ratio of r, which is 1.004^12 (= 1.049...), however, no part of the question seems to even suggest that the amount of withdrawal has to increase like this each year.

    I would much appreciate if someone can help me to understand this working on this mark scheme.

  2. #2
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    Thank you for your help, Denis

    Thank you for your help, Denis.

    I have found your explanation hard to follow and I do not understand especially the formula that you put:

    P = A*i / [1 - 1/(1 + i)^n]

    However, someone else has given me a clue and pointed out the fact that when you withdraw the 1000, you not only remove the money, you forfeit the interest you would have earned had you left it on deposit.

    So if r is the annual interest rate to figure out the amount in the account, the balance is:

    Year 1: 15000r^1 - 1000
    Year 2: (15000r^1 - 1000)r - 1000 = 15000r^2 - (1000r + 1000)
    Year 3: [15000r^2 - (1000r + 1000)]r - 1000 = 15000r^3 - (1000r^2 + 1000r + 1000)
    Year 4: [15000r^3 - (1000r^2 + 1000r + 1000)]r - 1000 = 15000r^4 - (1000r^3 + 1000r^2 +1000r + 1000)

    As you can see from above, there is a pattern in the sequence so you can generalise it as follows:

    Year n: 15000 r^n - (sum of geometric series of the first term of 1000 and the common ration of r)


    Quote Originally Posted by Denis View Post
    The amount of withdrawal does NOT increase.
    The "account" will be exactly like this:
    Code:
    YEAR  PAYMENT  INTEREST   BALANCE
      0                      15000.00
      1  -1000.00   736.05   14736.05 : 15000 * .04907 = 736.05
      2  -1000.00   723.10   14459.15
      3  -1000.00   709.51   14168.66
    ...
     25  -1000.00   166.65    2562.87
     26  -1000.00   125.76    1688.63
     27  -1000.00    82.86     771.49
     28   -809.35    37.86        .00
    A = 15000
    P = 1000
    i = .04907
    n = ?
    Formula is:
    P = A*i / [1 - 1/(1 + i)^n]
    Simplifies to:
    (1 + i)^n = P / (P - A*i)
    So:
    n = LOG[P / (P - A*i)] / LOG[1 + i] = 27.8056...

  3. #3
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    Quote Originally Posted by Masaru View Post
    As soon as Mary was 18 she decided to invest $15000 of thismoney in an account of the same type earning 0.4% interest per month. She withdraws $1000 every year on her birthday to buy herself a present. Determine how long it will take until there is no money in the account.

    The mark scheme says:
    r = 1.004^12 (= 1.049...)
    15000r^n - [1000(r^n-1)]/(r-1) = 0
    n = 27.8...
    So n = 28
    The others who have answered don't seem to have dealt with your actual question (though they are probably right about this not being the easiest way). What you say here,
    However, someone else has given me a clue and pointed out the fact that when you withdraw the 1000, you not only remove the money, you forfeit the interest you would have earned had you left it on deposit.

    So if r is the annual interest rate to figure out the amount in the account, the balance is:

    Year 1: 15000r^1 - 1000
    Year 2: (15000r^1 - 1000)r - 1000 = 15000r^2 - (1000r + 1000)
    Year 3: [15000r^2 - (1000r + 1000)]r - 1000 = 15000r^3 - (1000r^2 + 1000r + 1000)
    Year 4: [15000r^3 - (1000r^2 + 1000r + 1000)]r - 1000 = 15000r^4 - (1000r^3 + 1000r^2 +1000r + 1000)

    As you can see from above, there is a pattern in the sequence so you can generalise it as follows:

    Year n: 15000 r^n - (sum of geometric series of the first term of 1000 and the common ration of r)
    is a correct explanation of what the mark scheme said.

    Are you satisfied with this as the answer to your question?

    Quote Originally Posted by Masaru View Post
    However, I do not understand why the total amount forwithdrawal is "[1000(r^n-1)]/(r-1)" because the question says "She withdraws $1000 every year on her birthday...".So I do not understand why it is not simply"1000n" because it is a fixed amount of $1000 annually. Obviously "[1000(r^n-1)]/(r-1)" is the sum of geometric sequence with the first term of 1000 and common ratio of r, which is 1.004^12 (= 1.049...), however, no part of the question seems to even suggest that the amount of withdrawal has to increase like this each year.

    I would much appreciate if someone can help me to understand this working on this mark scheme.
    The amount you ask about, "[1000(r^n-1)]/(r-1)", is not the total amount withdrawn; it is the amount that must be deducted from the balance to account for both the withdrawals and the resulting lost interest.

    What I don't understand is how they solved their horrible equation for n with almost no explanation. Are you okay with that step?

  4. #4
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    Thank you for your concern, Dr. Peterson.

    Quote Originally Posted by Dr.Peterson View Post

    What I don't understand is how they solved their horrible equation for n with almost no explanation. Are you okay with that step?
    Yes, I should be OK because we are allowed to use graphing calculator for this question.

  5. #5
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    Quote Originally Posted by Masaru View Post
    Yes, I should be OK because we are allowed to use graphing calculator for this question.
    Excellent. That's what I suspected.

    Of course, the other way you were shown (using a formula you evidently have not learned) makes it possible to solve without that. But most real-life problems can't be solved by direct algebraic means anyway.

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