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    Hi

    A girl invests 500 at the start of each year in an account paying 3.5% interest per year. At the start of each year she puts in another 500. After how many years has this account accumulated more than 20000?

    Exam solutionz has a different approach but I took a different approach because I didn't understand his approach.

    So, i sed the account starts with 500 and the multiplicative factor of 3.5% will be 1.035 so that will be r

    so a = 500
    r = 1.035

    I then used my formula for Summing Gps and ended up with n = log2.4/log1.035
    = 25.4486

    Exam Solutionz ended up with n being greater than or equal to 24.86...
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    (Original post by NinCheng)
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    Quick skim, how does your method account for the extra 500 she adds in every year?
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    (Original post by Zacken)
    Quick skim, how does your method account for the extra 500 she adds in every year?
    idk but I got 25.44 was that by luck or
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    (Original post by NinCheng)
    idk but I got 25.44 was that by luck or
    Could you post a picture of the question or link to the examsolutions video or something?
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    (Original post by Zacken)
    Could you post a picture of the question or link to the examsolutions video or something?
    http://www.examsolutions.net/maths-r.../example-3.php
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    So your answer isn't correct. You need to write down the first few terms of the sequence and see if you notice something.

    1st year: 500(1.05)
    2nd year: (500 + 500(1.05))(1.05) = 500(1.05) + 500(1.05)^2
    3rd year: 500(1.035) + 500(1.035)^2 + 500(1.035)^3

    So nth year: 500(1.035 + 1.035^2 + ... + 1.035^n)

    So you want to find n s.t 500(1.0.35 + 1.035^2 + ... + 1.035^n) > 20000

    The inside of the bracket is a sum you can use your formula for.
 
 
 
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