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There are two positive numbers that can be inserted between 3 and 9 such that the first three are in geometric progression while the last three are in arithmetic progression. Find the sum of those two numbers.

  • math -

    let the sequence be
    3 a b 9

    then b/a = 9/b and b-a = 9-b
    from the first...
    b^2 = 9a
    a = b^2/9

    from the second...
    a = 2b-9

    so b^2/9 = 2b-9
    b^2 = 18b - 81
    b^2 - 18b + 81 = 0
    (b-9)(b-9) = 0
    b = 9
    then a = 9

    sum of these two numbers is 18

  • math -

    Thanks, this helped a lot! :)

  • ERROR - math -

    My solution is wrong, I misread the question.

    should be
    b/a = a/3 ---- a^2 = 3b

    b-a = 9-b ---> a = 2b-9

    then (2b-9)^2 = 3b
    4b^2 - 36b + 81 = 3b
    4b^2 -39b + 81 = 0
    b = (39 ± √225)/8
    = 27/4 or 3

    if b=3, then a = 3, sum of those two is 6
    if b = 27/4, then a = 9/2 or 4.5 , their sum is

    however.... 3 3 3 9 does not satisfy the original condition, while

    3 , 9/2 , 27/4 , 9 does work

    so the 2 numbers inserted are 9/2 and 27/4
    or 4.5 and 6.75
    their sum is 11.25

    sorry about the previous post.

  • math -

    Ah, okay I see. Thanks for correcting it.

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