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Math Help - Arithmetic and Geometric Progressions

  1. #1
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    Arithmetic and Geometric Progressions

    Positive numbers a1, a2, a3 are in arithmetic progression, while positive numbers g1, g2, g3 are in geometric progression. Given that a1 + g1 = 26, a2 + g2 = 50, a3 + g3 = 114, and a1 + a2 + a3 = 60, find both progressions.

    ***Please note the 1, 2 and 3 are meant to be subscripts
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  2. #2
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    What is your plan?

    Arithmetic Progression: a2 - a1 = a3 - a2
    Geometric Progression: g2/g1 = g3/g2

    Just a few hints. There are more.
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  3. #3
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    let, a1= a; a2 = a+d; a3 = a+2d and g1 = x; g2= x r; g3 = x r^2
    d: different and r: rasio.
    substitute the equality to the problem.
    ex:
    a1 + g1 = a + x = 36... (1)
    .
    .
    .
    etc

    hope it'll help ^_^
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  4. #4
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    Quote Originally Posted by chris520 View Post
    a1 + g1 = 26, a2 + g2 = 50, a3 + g3 = 114, and a1 + a2 + a3 = 60.....
    From these, can you "see" that g1 + g2 + g3 = 130?
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  5. #5
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    Quote Originally Posted by chris520 View Post
    Positive numbers a1, a2, a3 are in arithmetic progression, while positive numbers g1, g2, g3 are in geometric progression. Given that a1 + g1 = 26, a2 + g2 = 50, a3 + g3 = 114, and a1 + a2 + a3 = 60, find both progressions.

    ***Please note the 1, 2 and 3 are meant to be subscripts
    Let the Arithmetic progression be
    a2-d, a2, a2+d
    so that (3)a2=60 so a2=20.

    Then g2=30.

    Let the Geometric progression be
    (g2)/r, g2, r(g2)

    Then a1+a2+a3+g1+g2+g3=190=60+130
    so g1+g2+g3=130

    (g2)/r+g2+(r)g2=130
    30+30r+30r^2=130r
    30r^2-100r+30=0
    3r^2-10r+3=0
    (3r-1)(r-3)=0

    allows you to solve for "r" and write out the Geometric sequence.

    You can calculate the terms of the Arithmetic sequence from those.
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