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Math Help - Engineering economics homework help

  1. #1
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    Engineering economics homework help

    Hello everyone,

    Im looking to get some help with a few questions from an engineering economics course. In particular im hung up on the following question:

    Lenny's parents have decided to deposit money every month a bank account with interest rate of 10.39% per month. Lenny can take the money out as a reward for completing his university degree. Suppose Lenny will complete his degree 22 years from today. The parents will proceed to deposit 686.76 dollars at the end of month one and then will increase the deposit amount by 174.27 until Lenny's graduation. How much money will be deposited by Lenny's parents at the end of month 32?

    1) is it true that the amount im looking for does not include interest?
    2) i know this is a arithmetic gradient because of the constant increase each month, but what im not sure of is how to calculate the amount of the gradient because of question 1. If this does not include interest then the gradient formula i have is undefined since the denominator would be 0.

    I used the following formula:
    (p/g,i,n) = (1+i)^n -in -1 / (i^2)(1+i)^n

    where n is the number of peroids
    i is the interest rate

    Thank you for your help.
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  2. #2
    MHF Contributor
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    Your problem quite confusing:
    10.39% per month is over 120% per year!
    Stated is Lenny gets degree in 22 years; then in 32 months?

    Anyway, if question is SIMPLY: how much did the parents deposit?
    then an interest rate is not necessary.

    Looks to me that answer is:
    a=686.76, b=174.27
    Total deposited = (a+0b) + (a+1b) + (a+2b) + (a+3b) +.....+ (a+30b) + (a+31b).
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  3. #3
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    i was afraid of that. is there any way to do this without doing the same calculation 32 times?
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  4. #4
    MHF Contributor
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    686.76(32) + 174.27(31)(32)/2 = 108,414.24

    a = 686.76, b = 174.27, n = 32

    a(n) + b(n)(n-1)/2

    And if that friendly Banker chips in 10.39% each month: $461,585.58 !!
    Happy Lenny...
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