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Math Help - Geometric series

  1. #1
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    Geometric series

    A mortgage is taken out for 80,000$. It is to be paid by annual installments of 5000$ with the first payment being made at the end of the first year that the mortgage was taken out. Interest of 4% is then charged on any outstanding debt. Find the total time taken to pay off the mortgage.

    The answer given is 25 years.

    Can someone please help me with this problem?
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  2. #2
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    Lexington, MA (USA)
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    Hello, Seirin!

    A mortgage is taken out for $80,000.
    It is to be paid by annual installments of $5000 with the first payment
    being made at the end of the first year that the mortgage was taken out.
    Interest of 4% is then charged on any outstanding debt.
    Find the total time taken to pay off the mortgage.

    The answer given is 25 years. . . . . I don't agree
    This is an Amortization . . .

    The formula is: . A \;=\;P\frac{i(1+i)^n}{(1+i)^n-1}

    . . where: . \begin{Bmatrix}A &=& \text{periodic payment} \\<br />
P &=& \text{principal amount of loan} \\ i &=& \text{interest rate per period} \\ n &=& \text{number of periods} \end{Bmatrix}



    We have: . A = 5000,\;P = 80,000,\;i = 0.04

    Then: . 5000 \;=\;\frac{80,000(0.04)(1.04^n)}{(1.04)^n-1} \quad\Rightarrow\quad 5000 \;=\;\frac{3200(1.04)^n}{(1.04)^n-1} \quad\Rightarrow\quad 25 \;=\;\frac{16(1.04)^n}{(1.04)^n - 1}

    . . 25(1.04)^n - 25 \;=\;16(1.04)^n \quad\Rightarrow\quad 9(1.04)^n \;=\;25 \quad\Rightarrow\quad (1.04)^n \;=\;\frac{25}{9}

    Take logs: . \ln(1.04)^n \;=\;\ln\left(\frac{25}{9}\right) \quad\Rightarrow\quad n\!\cdot\!\ln(1.04) \:=\:\ln\left(\frac{25}{9}\right)


    Therefore: . n \;=\;\frac{\ln(\frac{25}{9})}{\ln(1.04)} \;=\;26.04876774 \;\approx\;\boxed{26\text{ years}}

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  3. #3
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    Sep 2008
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    Thanks Soroban

    But is there a way to solve this using geometric series formula? Coz this is a question from the geometric series chapter.
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