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Math Help - Geometric sequence

  1. #1
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    Geometric sequence

    Hello all!

    I have a problem: whenever I am doing word problems involving geometric or arithmetic sequences or series, I always end up in a pair of simultaneous equations that I do not know how to solve. For instance, I am currently doing the following problem

    Problem. The difference of the fifth term and the second term of a geometric series is equal to 156, and the the difference of the seventh term and the fourth term is equal to 1404. Find the first term of the geometric series and the common ratio.

    Ok, so I start by writing a pair of equations.

    Step 1.

    u_5-u_2=156\ (1) and u_7-u_4=1404\ (2),
    and then I plug in the standard definition of the [tex]nth[tex] terms of a geometric series u_n=aq^{n-1}, where a is the first term and q is the common ratio.

    I get the following set of equations

    aq^4-aq=156 and aq^6-aq^2=1404, but I do not know any technique to solve such a pair of simultaneous equations. I tried adding them together, solving for a, and other things, but I always end up in really complicated equations that I most certainly am not able to solve.

    Step 2.
    I notice that in both cases the difference is equal to two terms.
    In the case of eq. (1) the difference is equal to u_3+u_4=156, and in the case of eq. (2) the difference is equal to u_5+u_6=1404.
    So then I let u_3=b=aq^2 and I get the following equations


    \frac{b(1-q^2)}{1-q}=25 and \frac{bq^2(1-q^2)}{1-q}=1404
    So then I divide one by the other to get
    q^2=9 and q=\pm 3

    Step 3.
     <br />
b+bq=156<br />
4b=156
    b=39

    since b=aq^2
     <br />
a=\frac{13}{3}<br />
    But according to my book this answer is wrong.
    There is something wrong with my method, right? Is there a better way to do this problem?
    Thanks to anyone who replies.
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  2. #2
    MHF Contributor red_dog's Avatar
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    \left\{\begin{array}{ll}aq^4-aq=156\\aq^6-1q^3=1404\end{array}\right.\Rightarrow\left\{\begi  n{array}{ll}aq(q^3-1)=156\\aq^3(q^3-1)=1404\end{array}\right.

    Then \frac{aq(q^3-1)}{aq^3(q^3-1)}=\frac{156}{1404}\Rightarrow\frac{1}{q^2}=\frac  {1}{9}\Rightarrow q=\pm 3

    Can you continue?
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