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Math Help - Question on geometric progression

  1. #1
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    Question on geometric progression

    Hello! Please explain how do I solve this question! I am confused. X.x

    [A Levels N97/P1/15]

    A bank has an account for investors. Interest is added to the account at the end of each year at a fixed rate of 5% of the amount in the account at the beginning of that year. A man and woman both invest money

    (a) The man decides to invest $x at the beginning of one year and then a further $x at the beginning of the second and each subsequent year. He also decides that he will not draw any money out of the account, but just leave it, and any interest to build up.

    i) Show that, at the end of n years, when the interest for the last year has been added, he will have a total of $21[(1.05^n) - 1]x in his account.

    ii) After how many complete years will he have, for the first time, at least $12x in his account?

    b) The woman decides that, to assist her in her everyday expenses, she will withdraw the interest as soon as it has been added. She invests $y at the beginning of EACH year. Show that, at the end of n years, she will receive a total of $(1/40)[n(n+1)y] in interest.

    Thank you for helping!!
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  2. #2
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    Hello, Tangera!

    Here's part of it . . .


    A bank has an account for investors. Interest is added to the account
    at the end of each year at a fixed rate of 5% of the amount in the account
    at the beginning of that year. A man and woman both invest money

    (a) The man decides to invest $x at the beginning of one year
    and then a further $x at the beginning of each subsequent year.
    He also decides that he will not draw any money out of the account.

    i) Show that, at the end of n years, when the interest for the last year has
    been added, he will have a total of: . 21(1.05^n - 1)x dollars in his account.

    At the start of year 1, he invests x dollars.
    At the end of year 1, it is worth: .  1.05x dollars.

    At the start of year 2, he invests another x dollars ... total: 1.05x + x dollars.
    At the end of year 2, it is worth: . 1.05(1.05x + x) \:=\:1.05^2x + 1.05x dollars.

    At the start of year 3, he invests another x dollars ... total: 1.05^2x + 1.05x  + x dollars.
    At the end of year 3, it is worth: . 1.05(1.05^2x + 1.05x + x) \:=\:1.05^3x + 1.05^2x + 1.05x dollars.

    . . . and so on . . .

    At the end of year n, it is worth: . 1.05^nx + 1.05^{n-1}x + \hdots + 1.05^2x + 1.05x dollars.


    \text{The balance is: }\;S \;=\;1.05x\underbrace{\left(1 + 1.05 + 1.05^2 + \hdots + 1.05^{n-1}\right)}_{\text{geometric series}}

    . . The series has first term a = 1, common ratio r = 1.05, and n terms.

    . . Its sum is: . 1\cdot\frac{1.05^n-1}{1.05-1}\:=\:\frac{1.05^n-1}{0.05}

    Hence, we have: . S \;=\;1.05x\left(\frac{1.05^n-1}{0.05}\right) \quad\Rightarrow\quad S\;=\; 21(1.05^n-1)x



    ii) After how many complete years will he have at least \$12x in his account?
    When is S = 12x ?

    We have: . 21(1.05^n-1)x \;=\;12x \quad\Rightarrow\quad 1.05^n-1 \:=\:\frac{4}{7}\quad\Rightarrow\quad 1.05^n \:=\:\frac{11}{7}

    Take logs: . \log\left(1.05^n\right) \:=\:\log\left(\frac{11}{7}\right) \quad\Rightarrow\quad n\!\cdot\!\log(1.05) \:=\:\log\left(\frac{11}{7}\right)

    Therefore: . n \;=\;\frac{\log\left(\frac{11}{7}\right)}{\log(1.0  5)} \;=\;9.22638... \;\approx\;10\text{ years}


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