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    Geometric sequence

    The fifth term of a geometric series is 5! and the sixth term is 6!. What is the fourth term?

    Can someone post their answer and reasoning. I have my answer and think it's right but i really need e.c right now and want to make sure. Thanks!
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jarny View Post
    The fifth term of a geometric series is 5! and the sixth term is 6!. What is the fourth term?

    Can someone post their answer and reasoning. I have my answer and think it's right but i really need e.c right now and want to make sure. Thanks!
    For a geometric sequence {a_n}:
    a_n = a_1*r^{n - 1}
    where a_1 is the first term in the series and a_n is the nth term.

    So
    a_5 = 5! = 120 = a_1*r^4
    a_6 = 6! = 720 = a_1*r^5

    Two equations, two unknowns, we can solve this.

    Divide a_6 by a_5:

    a_6/a_5 = 6!/5! = 6 = (a_1*r^5)/(a_1*r^4) = r

    Thus r = 6.

    So
    120 = a_1*6^4

    Thus
    a_1 = 120/6^4 = 20/6^3 = 20/216 = 5/54

    So
    a_n = (5/54)*6^{n - 1}

    So
    a_4 = (5/54)*6^{4 - 1} = (5/54)*6^3 = (5/54)*216 = 20

    -Dan
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    got the same answer, thanks a lot
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    Hello, jarny!

    The fifth term of a geometric series is 5! and the sixth term is 6!.
    What is the fourth term?
    Dan's solution is absolutely correct.
    . . That's the approach I would have used.

    But upon reflection, I realized that I let the factorials dazzle me.
    . . There is a simpler solution.


    We are given: .a
    5 = 120 .and .a6 = 720

    Then: .r .= .a
    6/a5 .= .720/120 .= .6

    . . The "rule" is multiply-by-six.

    Therefore, the preceding term is: .a
    4 = 20.


    See? .We could have eyeballed the problem . . .

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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Soroban View Post

    . . There is a simpler solution.
    I always tell my students that I have a tendancy to make things harder than they have to be.

    -Dan
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