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Math Help - Applications of geometric series

  1. #1
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    Applications of geometric series

    A year plan is a national savings scheme requiring 12 monthly payments of a fixed amount of money on the same date each month. all savings earn intrest at a rate of x% per complete calendar months, his first payment has increased in value to:

    20 r^12 , where r =1+(x/100)

    show that the total value, after 12 complete calendar months, of all 12 payments is:

    20r(r^12 -1)/r-1

    hence calculate the toal interest received during the 12 months when the monthy rate of interest is 1/2%

    this is the question as it is given to me and its the final question on my assingment any help on this would be greatly apprecieted
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  2. #2
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    Quote Originally Posted by pmh118 View Post
    .... his first payment has increased in value to:
    20 r^12 , where r =1+(x/100)
    show that the total value, after 12 complete calendar months, of all 12 payments is:
    20r(r^12 -1)/r-1
    Clarification required...

    > 20 r^12 , where r =1+(x/100)
    This means deposits on 1st of month, 1st deposit immediately; is that your intent?

    > 20r(r^12 -1)/r-1
    Please check this. Is there a division by (r-1)? CLARIFY usinf brackets.
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  3. #3
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    Yes the deposit is on the 1st of each month

    and 20r(r^12 -1) all divided by (r -1)
    sorry if its messy im not sure howto enter equations in the forum
    Last edited by pmh118; October 12th 2010 at 07:54 AM. Reason: update text
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  4. #4
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    Quote Originally Posted by pmh118 View Post
    Yes the deposit is on the 1st of each month
    and 20r(r^12 -1) all divided by (r -1)
    OK; then your problem could be worded this way:

    20 is deposited on the 1st of each month over 1 year (so Jan 1st to Dec 1st).
    The annual interest rate is 12x% compounded monthly, hence x% per month.
    Therefore, as at Dec 31st, the first payment has increased in value to:
    20 r^12 , where r =1 + x/100.
    Show that the total value, as at Dec 31st, is:
    20r(r^12 -1) / (r-1).
    If the rate is 1/2% per month (hence r = 1.005), calculate the toal interest
    received during the 12 months.

    I'll give you a shorter example (hate typing!): 100 deposited for 6 months
    (Jan 1 to Jun 1) at a rate of 1% per month (hence r = 1.01):
    Code:
    Date    Deposit  Interest  Balance
    Jan.1    100.00       .00   100.00
    Feb.1    100.00      1.00   201.00
    Mar.1    100.00      2.01   303.01
    Apr.1    100.00      3.03   406.04
    May.1    100.00      4.06   510.10
    Jun.1    100.00      5.10   615.20
    Jun.30      .00      6.15   621.35
    
    S       =               100(1.01^6) + 100(1.01^5) + 100(1.01^4) + 100(1.01^3) + 100(1.01^2) + 100(1.01^1)
    S(1.01) = 100(1.01^7) + 100(1.01^6) + 100(1.01^5) + 100(1.01^4) + 100(1.01^3) + 100(1.01^2)
    
    S(1.01) - S = 100(1.01^7) - 100(1.01^1)
    S(1.01 - 1) = 100(1.01^7 - 1.01^1)
    
    S = 100(1.01)(1.01^6 - 1) / (1.01 - 1) = 621.35
    Above shows you how to "explain" the formula.
    As far as the total interest paid, it's simply:
    Final balance - total deposits; in my example: 621.35 - 100*6 = 21.35

    Hope that's enough to make your day!
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