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If 192;x;y;3 form a geometric progression, calculate the value of x and y.

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  1. 3d = 3-192 = -189
    d = -63

    192;129;66;3

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  2. 192,x,y,3,.....
    t1 ,t2,t3,t4,...
    if it is in GP then
    (t2/t1)=(t3/t2)=(t4/t3)
    then (t2/t1)=x/192.........(1)
    (t3/t2)=y/x...........(2)
    (t4/t3)=3/y...........(3)
    Now (1)=(2)

    x/192=y/x
    =>192y=x^2.......(4)

    Also
    (2)=(3)

    y/x=3/y

    =>3x=y^2.........(5)

    from (2)=> (y^2)/3 substitute in (4)

    =>192y=((y^2)/3)^2
    =>192y=(y^4)/9
    =>192*9=y^3
    =>1728=y^3

    =>y=12 substitute in (5)

    =>we get x=48

    therfore x=48,y=12
    then verify the series
    192,48,12,3........

    Hence solved.

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