The su three numbers in g.p. is 21 and the sum of their squares is 189 find the numbers.

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  1. Sum=21=A + Ar + Ar^2

    sumsquares: 189=A^2+(Ar)^2 + (Ar)^4


    squaring the first equation:
    21^2=A^2 (1+r+r^2)^2
    441=A^2 ( (1 +2r+3r^2+2r^3+r^4)
    441=A^2 (1+r^2+r^4)+A^2(2r+2r^2+2r^3)
    solve for A. Now go back to the first equation and solve for r.

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  2. A(1+r+r^2) = 21
    A(r^3-1)/(r-1) = 21
    A^2 (r^3-1)^2/(r-1)^2 = 441

    A^2 (1+r^2+r^4) = 189
    A^2 (r^6-1)/(r^2-1) = 441

    Now divide and you get to cancel a lot of factors, winding up with

    (r^2+r+1)/(r^2-r+1) = 441/189

    cross-multiply and clean things up, and you end with

    2r^2 - 5r + 2 = 0
    (2r-1)(r-2) = 0
    r = 2 or 1/2

    A(r^3-1)/(r-1) = 21
    A(7) = 21
    A = 3


    A(-7/8)/(-1/2) = 21
    A(7/4) = 21
    A = 12




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