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Thread: Please Help

  1. #1
    Junior Member VDestinV's Avatar
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    Question Please Help

    I posted a question last night but all that happened was I became even more confused.

    The homework question is:

    Find the sum of each geometric series.
    a) S10 in the series $2500 + $2562.50 + $2626.56 + ...


    now i would normally use the formula

    tn= ar^n-1 except I do not know what tn equals. To find the ^n I need to know what tn is. Now can some one please help me.
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  2. #2
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    Hello, VDestinV!

    Find the sum of the geometric series.

    a) $\displaystyle S_{10}$ in the series: .$\displaystyle \$2500 + \$2562.50 + \$2626.56 + \hdots$
    First, find the common ratio $\displaystyle r.$


    We have: .$\displaystyle \frac{2562.50}{2500} \:=\:1.025$

    . . . .and: .$\displaystyle \frac{2626.56}{2562.50} \:=\:1.024999024$

    The common ratio appears to be: .$\displaystyle r \:=\:1.025$


    Then: .$\displaystyle S_{10} \;=\;2500\,\frac{1.1025^{10}-1}{1.025-1} \;=\;128,008.4544 \:\approx\:\$128,008.45 $

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  3. #3
    Junior Member VDestinV's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, VDestinV!

    First, find the common ratio $\displaystyle r.$


    We have: .$\displaystyle \frac{2562.50}{2500} \:=\:1.025$

    . . . .and: .$\displaystyle \frac{2626.56}{2562.50} \:=\:1.024999024$

    The common ratio appears to be: .$\displaystyle r \:=\:1.025$


    Then: .$\displaystyle S_{10} \;=\;2500\,\frac{1.1025^{10}-1}{1.025-1} \;=\;128,008.4544 \:\approx\:\$128,008.45 $
    wow thanks... but uhh what about the n?
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  4. #4
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    10 is n, as I explained in the other thread.
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  5. #5
    Junior Member VDestinV's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, VDestinV!

    First, find the common ratio $\displaystyle r.$


    We have: .$\displaystyle \frac{2562.50}{2500} \:=\:1.025$

    . . . .and: .$\displaystyle \frac{2626.56}{2562.50} \:=\:1.024999024$

    The common ratio appears to be: .$\displaystyle r \:=\:1.025$


    Then: .$\displaystyle S_{10} \;=\;2500\,\frac{1.1025^{10}-1}{1.025-1} \;=\;128,008.4544 \:\approx\:\$128,008.45 $
    I was looking over your calculations...

    shouldn't .$\displaystyle S_{10} \;=\;2500\,\frac{1.1025^{10}-1}{1.025-1} \;=\;128,008.4544 \:\approx\:\$128,008.45 $

    actually be .$\displaystyle S_{10} \;=\;2500\,\frac{1.025^{10}-1}{1.025-1} \;=\;28,008.4544 \:\approx\:\$28,008.45 $
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  6. #6
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    Yes, that's correct.
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  7. #7
    Junior Member VDestinV's Avatar
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    okay but i tried to do that with this one but it didnt work

    s12 in the series 60, 12, 2.4, ...
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  8. #8
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    Quote Originally Posted by VDestinV View Post
    okay but i tried to do that with this one but it didnt work

    s12 in the series 60, 12, 2.4, ...
    $\displaystyle a = 60$

    $\displaystyle r = \frac{12}{60} = 0.2$

    $\displaystyle n = 12$

    $\displaystyle S_{12} = \frac{60(1 - 0.2^{12})}{1 - 0.2}$

    $\displaystyle S_{12} = 75$
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