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Math Help - Arithmetical sequence

  1. #1
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    Arithmetical sequence

    I have been trying to solve this arithmetical sequence but I can't find a1;a2;a3 if:
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  2. #2
    TD!
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    Is it a_3 or a_5?

    Anway, use the basic property of an arithmetic sequence to convert everything in terms of a_1 and the difference of the sequence, I'll call that v.

    We then have: a_n  = a_1  + \left( {n - 1} \right)v

    Use this on all terms except a_1 of course and then you have a system of two equations in two unknowns (a_1 and v).
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  3. #3
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    It is a_3.
    I used the formula what you mentioned and I substituted v into the second operation. I got cube equation -7a_1^3-6a_1^2+288a_1=1000
    I am not able to find the a_1 from that cube equation.
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  4. #4
    TD!
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    So we have then

    \left\{ \begin{array}{l}<br />
 a_1  + a_2  + a_3  =  - 12 \\ <br />
 a_1  \cdot a_2  \cdot a_3  = 80 \\ <br />
 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}<br />
 a_1  + \left( {a_1  + v} \right) + \left( {a_1  + 2v} \right) =  - 12 \\ <br />
 a_1  \cdot \left( {a_1  + v} \right) \cdot \left( {a_1  + 2v} \right) = 80 \\ <br />
 \end{array} \right

    \left\{ \begin{array}{l}<br />
 3a_1  + 3v =  - 12 \\ <br />
 a_1 ^3  + 3a_1 ^2 v + 2a_1 v^2  = 80 \\ <br />
 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}<br />
 a_1  + v =  - 4 \\ <br />
 a_1 ^3  + 3a_1 ^2 v + 2a_1 v^2  = 80 \\ <br />
 \end{array} \right

    From the first equation we have that v = -4 -a_1, substitute this in the second equation, the cubic term will disappear:

    a_1 ^3  + 3a_1 ^2 \left( { - 4 - a_1 } \right) + 2a_1 \left( { - 4 - a_1 } \right)^2  = 80 \Leftrightarrow 4a^2  + 32a = 80

    Solving this should be doable
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  5. #5
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    Countinuing where TD! left of, we have that,
    a_2=a_1+k
    a_3=a_1+2k
    Where k is the common difference.
    Thus,
    a_1+a_2+a_3=-12
    Becomes,
    3a_1+3k=-12
    Thus,
    a_1+k=-4 (1)
    Now for the second equation,
    a_1(a_1+k)(a_1+2k)=80
    Thus,
    a_1(-4)(a_1+2k)=80 by equation (1)
    Thus,
    a_1(a_1+2k)=-20
    Simply (2) as,
    a_1(a_1+k+k)=-20
    Thus,
    a_1(k-4)=-20
    Thus,
    a_1k-4a_1=-20 (2)
    Now you have the two equations, solve them,
    substitute for a_1 from (1),
    (-4-k)k-4(-4-k)=-20
    Thus,
    -4k-k^2+16+4k=-20
    Simplify,
    k^2=36
    Thus,
    (k,a_1)=(6,-10)\mbox{ or }(-6,-14)
    Now just check which one works.
    Q.E.D.
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  6. #6
    TD!
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    Quote Originally Posted by ThePerfectHacker
    Thus,
    (k,a_1)=(6,-10)\mbox{ or }(-6,-14)
    Now just check which one works.
    Q.E.D.
    There's a slight mistake here, for k = -6 we have that a_1 = 2.

    Furthermore, there is not reason to check which one 'works' since there's no physical interpretation or whatever which has to be checked. All we did was solving the system of equations algebraiclly so both of our solutions are fine, i.e. meet the conditions which were given.
    Last edited by TD!; January 27th 2006 at 12:41 PM.
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  7. #7
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    Quote Originally Posted by TD!
    There's a slight mistake here, for k = -6 we have that a_1 = 2.

    Furthermore, there is not reason to check which one 'works' since there's no physical interpretation or whatever which has to be checked. All we did was solving the system of equations algebraiclly so both of our solutions are fine, i.e. meet the conditions which were given.
    I erred again, but it is the method not the answer that is important.
    You statement about the checking the two solutions; I know the two solutions work for the algebraic equations but not necessarily for the conditions given for the problem thus we must check.
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  8. #8
    TD!
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    Well the problem was that knowing we were dealing with an arithmetic sequence, "find a1;a2;a3 if: (the two equations)".
    Both solutions are fine, there is no reason to discard one since there's nothing left to check! Unless, of course, totalnewbie didn't give the entire problem (but I think he did or at least I assumed so).
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  9. #9
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    It seems to me that my math book contains a mistake.
    There is a_1+a_2+a_5=-12
    According to your answers I belive that this must be a_1+a_2+a_3=-12
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  10. #10
    TD!
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    That's why I asked if it was a_3 or a_5. From what I understand now, it was actually a_5 used in the equation but you were asked to find a_1,a_2 and a_3?

    In that case, just repeat the exact same method but use a_5 instead of a_3, a_5 can be converted to a_1 + 4v.

    You have the entire method now, try it
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