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Math Help - Arithmetic and geometric sequences

  1. #1
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    Arithmetic and geometric sequences

    How do you solve a) and b)

    Ray and Simon are using two different training programs to prepare for the rottnest marathon.

    Ray runs every day. In the first week he runs 5km each day and then increases this distance by 2km each week thereafter.

    Simon also runs every day. In the first week he runs 10 km and then increases this by 10% each week there after.

    a) How long does it take for simon to run at least the marathon distance of 42.km?

    b) How far would Rayy be running after 15 weeks of traning:
    (SOLVED)

    I know that b) it was a arithmetic sequence and a) involves a geometric sequence. I dont know how to solve a).

    Thank you
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  2. #2
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    Quote Originally Posted by atom360 View Post

    Ray runs every day. In the first week he runs 5km each day and then increases this distance by 2km each week thereafter.

    a) How long does it take for simon to run at least the marathon distance of 42.km?
    The sequence is \{5,5+2,5+2\times 2, 5+3\times 2, \dots\}

    Where a = 5 and d = 2

    S_n = \frac{n}{2}[a+(n-1)\times d]

    42 = \frac{n}{2}[5+(n-1)\times 2]

    Can you solve for n?
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  3. #3
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    Hello, atom360!

    Ray and Simon are using two different training programs to prepare for a marathon.

    Ray runs every day. In the first week he runs 5km each day
    and then increases this distance by 2km each week thereafter.

    Simon also runs every day. In the first week he runs 10 km
    and then increases this by 10% each week there after.

    a) How long does it take for Simon to run at least the marathon distance of 42 km?

    b) How far would Ray be running after 15 weeks of traning? solved

    I know that (a) involves a geometric sequence. . . . . Right!

    Simon's sequence has: . \begin{array}{ccccc}\text{first term:} & a &=& 10 \\ \text{common ratio:} &r &=& 1.1 \end{array}

    The sum of a geometric series is: . S_n \;=\;a\,\frac{r^n-1}{r-1}


    We have: . 10\,\frac{1.1^n - 1}{1.1-1} \:=\:42 \quad\Rightarrow\quad 1.1^n - 1 \:=\:0.42 \quad\Rightarrow\quad 1.1^n \:=\:1.42

    Take logs: . \ln(1.1)^n \:=\:\ln(1.42) \quad\Rightarrow\quad n\ln(1.1) \:=\:\ln(1.42)

    Therefore: . n \;=\;\frac{\ln(1.42)}{\ln(1.1)} \;=\;3.6789112476

    It will take Simon about 3.7 weeks to run a total of 42 kilometers.

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