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Math Help - Arithmetic Sequences

  1. #1
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    Arithmetic Sequences

    I have been having trouble with the following problem:
    Write the next term in the sequence. Then write a rule for the nth term.

    7,-8,9,-10... I found out that the next number in the sequence is 11, but I have yet to figure out the rule for the nth term. I have been trying to solve this problem for a while and cannot. Can you help?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by KittyOfDoom View Post
    I have been having trouble with the following problem:
    Write the next term in the sequence. Then write a rule for the nth term.

    7,-8,9,-10... I found out that the next number in the sequence is 11, but I have yet to figure out the rule for the nth term. I have been trying to solve this problem for a while and cannot. Can you help?
    a_n=(n+6)(-1)^{n+1}
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  3. #3
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    Hello, KittyOfDoom!

    Write the next term in the sequence. Then write a rule for the nth term.
    . . . 7,\:-8,\:9,\:-10,\:\hdots

    Ignoring the signs, we have: . 7,\:8,\:9,\;10,\:\hdots

    . . The rule (so far) is: . f(n) \:=\:n+6


    Now, how do we make the signs alternate?
    . . If you've never seen it done, this can be baffling.

    Consider (-1)^n . . . negative-one raised to consecutive powers.


    We get:
    . . . . . . . \begin{array}{ccc}(-1)^1 &=& -1 \\ (-1)^2 &=& +1 \\ (-1)^3 &=& -1 \\ (-1)^4 &=& +1 \\ (-1)^5 &=&-1 \\ \vdots & & \vdots \end{array}\quad\hdots . See? .Alternating signs!

    If we want to start with "minus", we use: . (-1)^n

    If we want to start with "plus", we use: . (-1)^{n+1}


    For this problem, the rule is: . \boxed{f(n) \;=\;(-1)^{n+1}(n+6)}

    . . Plug in n = 1,2,3,\hdots and check it out.

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  4. #4
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    Thank you for the help ^.^
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