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Math Help - Geometric progression

  1. #1
    Member Rimas's Avatar
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    Geometric progression

    if the geometric progression is 216 and their sum is 19 find the largest of the three numbers
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rimas View Post
    if the geometric progression is 216 and their sum is 19 find the largest of the three numbers
    A geometric series is defined as the numbers:
    a_{n+1} = r*a_n for n = 0, 1, 2, ....

    where some a_0 is known.

    As an explicit formula for the a_n this becomes:
    a_n = c*r^n
    where c is the constant c = a_0.

    So we have three numbers in geometric progression and their product is 216 and their sum is 19:
    [c*r^n]*[c*r^{n+1}]*[c*r^{n+2}] = 216
    c*r^n + c*r^{n+1} + c*r^{n+2} = 19

    We can, without loss of generality set n = 0.

    [c]*[c*r]*[c*r^2] = 216
    c + c*r + c*r^2 = 19

    or
    c^3*r^3 = 216 => c*r = 6 so we have that c = 6/r

    So
    c(1 + r + r^2) = 19

    (6/r)(1 + r + r^2) = 19

    6 + 6r + 6r^2 = 19r

    6r^2 - 13r + 6 = 0

    Which has solutions:
    r = 3/2 and r = 2/3

    Thus c = 6/r so c = 4 or c = 9 respectively.

    Thus the two possible series are:
    4, 6, 9 (4*6*9=216 and 4+6+9=19 check!)
    9, 6, 4 (9*6*4=216 and 9+6+4=19 check!)
    (Yes, these ARE two distinct series.)

    Either way, the largest number in the series is 9.

    -Dan
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