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Math Help - Just a quick one

  1. #1
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    Just a quick one

    What are the formulaes for:

    Arithmatic progression, if i know how many numbers there are, their sum, and their product?

    and

    Geometric progression, if i know the second term and the sum to infinity?
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  2. #2
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    Quote Originally Posted by Nixietube View Post
    What are the formulaes for:

    Arithmatic progression, if i know how many numbers there are, their sum, and their product? and Geometric progression, if i know the second term and the sum to infinity?
    " Arithmatic progression, if i know how many numbers there are, their sum?":
    First of all I think we need to know the first term and common difference.

    With that we could tell you the formula, but I think its better you consult your text book. Text books generally have a flow that makes it easy to understand, remember and apply.

    "and their product?"
    I am sorry, this could be hard. I have no idea of this.

    I think its better if you post a problem that demands the use of these formulae. Can you post one?

    "and Geometric progression, if i know the second term and the sum to infinity?"
    Afterthought: I think you have a problem at hand and thus you want to know the formulae. So could you post those exactly as in the question?
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  3. #3
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    It doesn't give the first number in the series, just the sum of the three numbers, and the product. Sorry i don't have the question at hand
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  4. #4
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    Are they consecutive numbers of the sequence ?

    Can you write all down ?
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  5. #5
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    The sum of 3 numbers in an arithmatic progression are 18 and their product is 120?

    I found some formulaes, but most require you to know the first term
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  6. #6
    Moo
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    Quote Originally Posted by Nixietube View Post
    The sum of 3 numbers in an arithmatic progression are 18 and their product is 120?

    I found some formulaes, but most require you to know the first term
    Assume that they are consecutive numbers... a ~,~b~,~c, in an arithmetic sequence of constant progression r.

    b=a+r and c=a+2r

    18=a+b+c=3a+3r \implies {\color{red}a+r}=6 \quad a=6-r

    120=abc=a({\color{red}a+r})(a+2r)=a*6*(6+r)=6*(6-r)(6+r)
    \implies 20=36-r^2 \implies r^2=16

    etc.



    ---------------------
    A slightly different way is to write :

    a=b-r, c=b+r


    18=a+b+c=b-r+b+b+r=3b \implies b=6

    120=abc=(6-r)6(6+r) \implies 20=36-r^2

    etc
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  7. #7
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    Thankyou!
    When i went through it i ended u with 2,6,10
    2+6+10 = 18
    2x6x10 =120

    Edit, just in case anyone else is doing something like this,
    I re-arranged
    20 = 36-r^2

    To give
    r^2 = 36 - 20

    Therefore
    r = 4

    6+4 = 10
    6-4 = 2
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  8. #8
    Moo
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    Therefore
    r = 4
    Uh-oh !
    It can be r=-4

    But it doesn't change anything ^^



    For your geometric progression, do you have an example ?
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  9. #9
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    I managed to stumble my way through the geometric progression one, but thankyou anyway moo, you've been a great help!
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