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Math Help - determining formula for sequences.

  1. #1
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    determining formula for sequences.

    Im kinda having trouble trying to determine the general forumla for these sequences.

    5,4,9,13,22

    and one for annual salary

    42 000, 43 680, 45 427.20, 47 244.29

    thanks in advance =]
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  2. #2
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    Quote Originally Posted by ferken View Post
    Im kinda having trouble trying to determine the general forumla for these sequences.

    5,4,9,13,22

    and one for annual salary

    42 000, 43 680, 45 427.20, 47 244.29

    thanks in advance =]
    For the first one, I'm not sure, but it works actually,

    Let the pth term be n
    so the next term will be n + (p-1)th term
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  3. #3
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    To get the general term for the annual salary you just have to take the product of the previous term with 1.04

    43 680 = 42 000(1.04)
    45 427.20 = 43 680(1.04)
    47 244.29 = 45 427.20(1.04)

    Hope you will be able to derive the general formula based on the help provided

    Quote Originally Posted by ferken View Post
    Im kinda having trouble trying to determine the general forumla for these sequences.

    5,4,9,13,22

    and one for annual salary

    42 000, 43 680, 45 427.20, 47 244.29

    thanks in advance =]
    Follow Math Help Forum on Facebook and Google+

  4. #4
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    Hello, ferken!

    Determine the general forumla for these sequences.

    (a)\;\;5, \:4, \:9, \:13, \:22, \hdots

    If you look, you see that each term is the sum of the preceding two terms.

    . . \begin{array}{ccc}5+4 &=& 9 \\ 4 + 9 &=& 13 \\ 9 + 13 &=& 22 \\ 13 + 22 &=& 35 \\ \vdots && \vdots \end{array}


    Therefore: . \boxed{a_n \:=\:a_{n-1} + a_{n-2} }\quad\text{with }a_1 = 5,\;a_2 = 4




    and one for annual salary

    (b)\;\; 42,\!000,\;43,\!680,\; 45,\!427.20,\; 47,\!244.29,\;\hdots
    This one is not an "additive" sequence . . . try ratios.

    . . \begin{array}{ccc}\dfrac{43,\!680}{42,\!000} &=& 1.04 \\ \\[-2mm]<br />
\dfrac{45,\!427.20}{43,\!680} &=& 1.04 \\ \\[-2mm]<br />
\dfrac{47,\!244.29}{45,\!427.20} & \approx & 1.04 \end{array}


    Each term is 1.04 times the preceding term . . .

    . . \boxed{a_n \;=\;1.04a_{n-1}}


    It is a geometric sequence with: first term a = 42,\!000 and common ratio r = 1.04

    . . \boxed{a_n \;=\;42,\!000(1.04)^{n-1}}

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