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    maths C2 chapter geometric sequences and series.

    Question is:I invest £A in the bank at a rate of interest of 3.5% per annum how long will it be before i double my money
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    (Original post by ballerguids)
    maths C2 chapter geometric sequences and series.

    Question is:I invest £A in the bank at a rate of interest of 3.5% per annum how long will it be before i double my money
    Use the Sum to n formula
    You are given that a=A and Sn = 2A and you know r so just need to solve what n is (prob using logs)
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    (Original post by ballerguids)
    maths C2 chapter geometric sequences and series.

    Question is:I invest £A in the bank at a rate of interest of 3.5% per annum how long will it be before i double my money
    Use Sn=\dfrac{a(1-r^n)}{1-r}
    Where Sn=2A
    a=A
    and r=1.035
    Then expand, simplify and solve using logs.
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    I have a similar question..

    Could someone help me with a Geometric and Sequences question?

    2nd term = 16
    Sum to infinity = 100

    Work out the 2 possible values of the common ration and the corresponding first terms.

    Thank you!


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    (Original post by Annabel_lear)
    I have a similar question..

    Could someone help me with a Geometric and Sequences question?

    2nd term = 16
    Sum to infinity = 100

    Work out the 2 possible values of the common ration and the corresponding first terms.

    Thank you!


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    A geometric sequence takes the following form

    a, ar, ar^2,.....,ar^{n-1}

    where a is the first term and r is the common ratio.

    The sum to infinity is given by

    \displaystyle S_{\infty} = \dfrac{a}{1-r}

    valid for -1 < r < 1, denoted by |r|<1, so you can check your answer
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    (Original post by Indeterminate)
    A geometric sequence takes the following form

    a, ar, ar^2,.....,ar^{n-1}

    where a is the first term and r is the common ratio.

    The sum to infinity is given by

    \displaystyle S_{\infty} = \dfrac{a}{1-r}

    valid for -1 < r < 1, denoted by |r|<1, so you can check your answer
    I understand that ar=16 and a/1-r = 100
    and that you have to solve them by simultaneous equations but I'm not sure how
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    (Original post by Annabel_lear)
    I understand that ar=16 and a/1-r = 100
    and that you have to solve them by simultaneous equations but I'm not sure how
    ar=16 \Rightarrow a=....

    fill that in and substitute into the other equation
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    (Original post by Indeterminate)
    ar=16 \Rightarrow a=....

    fill that in and substitute into the other equation
    I changed ar=16 to a= 16/r
    so if you put that into the other equations its
    16/r /1-r =100
    That's where I get stuck!
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    (Original post by Annabel_lear)
    I changed ar=16 to a= 16/r
    so if you put that into the other equations its
    16/r /1-r =100
    That's where I get stuck!
    \dfrac{\frac{16}{r}}{1-r} = 100

    Hint:

    \dfrac{(\frac{x}{2})}{2} = \dfrac{x}{2\times 2} = \dfrac{x}{4}
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    (Original post by Indeterminate)
    \dfrac{\frac{16}{r}}{1-r} = 100

    Hint:

    \dfrac{(\frac{x}{2})}{2} = \dfrac{x}{2\times 2} = \dfrac{x}{4}
    So does it turn to 16/1-r2 = 100
    then
    16 = 100 - 100r2 ​?
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    (Original post by Annabel_lear)
    So does it turn to 16/1-r2 = 100
    then
    16 = 100 - 100r2 ​?
    Not quite

    Notice that you are multiplying by \frac{r}{r}, which is 1, so you're not changing anything.

    So

    \dfrac{\frac{16}{r}}{1-r} = 100 \Rightarrow \dfrac{16}{r(1-r)} = 100

    Make sense?
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    (Original post by Annabel_lear)
    I understand that ar=16 and a/1-r = 100
    and that you have to solve them by simultaneous equations but I'm not sure how
    Personally I would start with the 2nd equation which can be rearranged to give a = 100(1-r) and then plug this into the first equation, which gives you:

    100r(1-r) = 16

    You can then expand this and rearrange to get a quadratic in r.
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    (Original post by Indeterminate)
    Not quite

    Notice that you are multiplying by \frac{r}{r}, which is 1, so you're not changing anything.

    So

    \dfrac{\frac{16}{r}}{1-r} = 100 \Rightarrow \dfrac{16}{r(1-r)} = 100

    Make sense?
    Ohhhhh Yupp! Thank you


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