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Math Help - Sequences and series

  1. #1
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    Sequences and series, a financial application with interest rates, please help me

    Dear forum members, could you please verify that my calculations are correct



    A person takes a two million euro loan from a bank with an annual fixed interest of 6%. He plans to pay it back in 10 years. In order to be able to complete his goal, how much must he pay per month?


    q=1,005 (the coefficient that will tell us the interest added to the loan)

    n=10 (the years we want to spend on paying the loan)

    a= 2 000 000 (the amount of money taken)

    b= ? (the amount of money we have to pay per month)



    q^n*a -\frac{b(1-q)}{1-q}\leq0


    Oh yeah, and if anyone was wondering I used geometric sum for the fraction part.


    I got for my answer for solving b, about 20 5541.1457 , but I highly doubt this is correct.


    So if someone could check it, I would be very grateful.
    Last edited by Coach; March 25th 2008 at 11:59 AM.
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  2. #2
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    Hello, Coach!

    Sorry, your answer is incorrect . . .


    A person takes a $2,000,000 loan from a bank with an annual fixed interest of 6%.
    He plans to pay it back in 10 years.
    In order to be able to complete his goal, how much must he pay per month?
    There is an Amortization Formula: . A \;=\;P\frac{i(1+i)^n}{(1+i)^n-1}
    . . where: . \begin{Bmatrix}A & = &\text{periodic payment} \\ P &=& \text{principal} \\ i &=& \text{periodic interest rate} \\ n&=&\text{number of periods} \end{Bmatrix}


    We have: . P \:=\:2,000,000,\;i \:=\:\frac{6\%}{12} \:=\:0.005,\;n \:=\:120

    Hence: . A \;=\;2,000,000\cdot\frac{0.005\cdot1.005^{120}}{(1  .005)^{120}-1}  \;=\;22,204.10039


    Therefore, his monthly payment will be: . \$22,204.10


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    The derivation of the Formula is lengthy but quite elementary.

    I think I can recreate the steps . . . if anyone is interested.

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, Coach!

    Sorry, your answer is incorrect . . .


    There is an Amortization Formula: . A \;=\;P\frac{i(1+i)^n}{(1+i)^n-1}
    . . where: . \begin{Bmatrix}A & = &\text{periodic payment} \\ P &=& \text{principal} \\ i &=& \text{periodic interest rate} \\ n&=&\text{number of periods} \end{Bmatrix}


    We have: . P \:=\:2,000,000,\;i \:=\:\frac{6\%}{12} \:=\:0.005,\;n \:=\:120

    Hence: . A \;=\;2,000,000\cdot\frac{0.005\cdot1.005^{120}}{(1  .005)^{120}-1}  \;=\;22,204.10039


    Therefore, his monthly payment will be: . \$22,204.10


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    The derivation of the Formula is lengthy but quite elementary.

    I think I can recreate the steps . . . if anyone is interested.






    Oh, thank you so much!!!!!!!!!!!!!


    Well, I almost got it
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  4. #4
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    However, if you could please explain why my approach did not work?
    Is it wrong to make it an inequality, and use the formulas for solving a geometric sum?
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  5. #5
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    And when I try checking it


    I get that the final amount with interets added that he would have to pay would be 1,05^{10}*2000 000 and that does not equal the monthly payment times 120?


    Why is that? Shouldn't they equal each other.
    Last edited by Coach; March 25th 2008 at 01:09 PM.
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