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Math Help - Application of Ap/GP

  1. #1
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    Application of Ap/GP

    i already try for question no Q1 as below, but the answer is not same from text book. so i hope some body can help me here.

    Q1. A man deposits RM 500 in a bank. The interest receive of 5% per year is kept in bank. Find
    (a) the number of years it will take for the total savings to exceed RM750.
    (the ans: 9 but my ans 11)

    (b) the total saving after 8 years.

    Q2. On 1 st January each year, a man puts RM 50 in a bank whick give 4% interest each year. What is the value of his saving on 31 st December of the tenth year?

    Tq for your time
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  2. #2
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    Q1. 1st year: 500\cdot 1.05, 2nd year: (500\cdot 1.05)\cdot 1.05, etc. So nth year: 500\cdot (1.05)^n, therefore this should hold: 500\cdot (1.05)^n\ge 750\leftrightarrow (1.05)^n\ge 1.5. Take the logarithm of both sides (base 1.05, you can without reversing the sign, because it is greater than 1): \log_{1.05}(1.05)^n\ge \log_{1.05}1.5 by definition it is: n\ge\log_{1.05}1.5\approx 8.31039. So the first integer after...
    Q2. Using the same idea: end of 10th year: 50\cdot (1.04)^{10}=1446.27
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  3. #3
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    Quote Originally Posted by james_bond View Post
    Q1. 1st year: 500\cdot 1.05, 2nd year: (500\cdot 1.05)\cdot 1.05, etc. So nth year: 500\cdot (1.05)^n, therefore this should hold: 500\cdot (1.05)^n\ge 750\leftrightarrow (1.05)^n\ge 1.5. Take the logarithm of both sides (base 1.05, you can without reversing the sign, because it is greater than 1): \log_{1.05}(1.05)^n\ge \log_{1.05}1.5 by definition it is: n\ge\log_{1.05}1.5\approx 8.31039. So the first integer after...
    Q2. Using the same idea: end of 10th year: 50\cdot (1.04)^{10}=1446.27
    i will try to calculate it back..it consider under G.P?
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  4. #4
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    It is a geometric progression. (Not sure what you're asking.)
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  5. #5
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    Quote Originally Posted by james_bond View Post
    It is a geometric progression. (Not sure what you're asking.)
    yes. it what i mean.

    ans
    Q1
    (a) 9
    (b) rm 738.70

    Q2
    rm 624.30
    Last edited by nikk; January 30th 2009 at 10:36 PM. Reason: solution
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  6. #6
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    Hi, sorry for not replying sooner.
    Q1(b): n=8, so: 500\cdot (1.05)^8\approx 738.728
    Q2: sorry I misread and misunderstood the question and wrote some stupid ****. Sorry for that.
    1st year (31st december) he has: 50\cdot 1.04, next year he has: 50\cdot 1.04^2+50\cdot 1.04, etc. 10th year: 50\cdot\left(1.04^{10}+1.04^9+\ldots +1.04\right)=50\cdot\frac{1\cdot(1.04^{10+1}-1.04^1)}{1.04-1}\approx 624.318 .
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