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Math Help - [SOLVED] Sequence Question

  1. #1
    Member looi76's Avatar
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    [SOLVED] Sequence Question

    Question:
    Each year a company gives a grant to a charity. The amount given each year increases by 5\% of its value in the preceding year. The grant in 2001 was \$5000. Find:

    (i) the grant given in 2011,
    (ii) the total amount of money given to the charity during the years 2001 to 2011 inclusive.


    Attempt:

    a = \$5000
    r = 5\% = 0.05

    2001 \rightarrow 2011 = 10 years

    U_{11} = a.r^{10}
    U_{11} = 5000 \times 0.05^{10} = 4.88^{-10}

    I don't think that is possible ! How can I solve this question?
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  2. #2
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    Maybe r=1.05
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  3. #3
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    Hello,

    r is not 5%, but 1.05, because you keep the previous amount.
    Let's take an example :

    The first year, they give 5000$.
    The following year, they will give 5000 increased of 5%.
    This means that they will give 5000+5%*5000=5000+.05*5000=5000*1.05

    The following year, they will give 5000*1.05 added up by 5%.
    This means that they will give 5000*1.05+5%*(5000*1.05)=(5000*1.05)*1.05=5000*1.0 5

    Is it clear for you now ?
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  4. #4
    Member looi76's Avatar
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    Quote Originally Posted by looi76 View Post
    (ii) the total amount of money given to the charity during the years 2001 to 2011 inclusive.
    S_n = \frac{a(1 - r^n)}{1 - r}

    S_n = \frac{5000(1 - 1.05^{10})}{1 - 1.05}

    S_n = \$62889.46

    Where did I go wrong? Answer in text book is $71034
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  5. #5
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    Quote Originally Posted by looi76 View Post
    S_n = \frac{a(1 - r^n)}{1 - r}

    S_n = \frac{5000(1 - 1.05^{10})}{1 - 1.05}

    S_n = \$62889.46

    Where did I go wrong? Answer in text book is $71034
    Hi looi76,

    When the question says inclusive they mean both the year 2001 and 2011 should be included. So actually there are 11 years in all. So your n=11...

    S_{11} = \frac{5000(1 - 1.05^{11})}{1 - 1.05} = \$71033.93
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