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1)Find the sum of the first eight terms of the Geometric progression 256,128,64,32

2)How many terms should be taken from the Geometric progression 4,12,36 for the sum to be 2188

  • math - ,

    1)
    The sum of n numbers in Geometric progression is:

    Sn=a1*[(1-q^n)/(1-q)]

    Where:
    a1 is first number in progresion
    q is the common ratio.

    In your case:

    a1=32
    q=2

    Sn=S8=32*[(1-2^8)/81-2]

    S8=32*[(1-256)/(1-2)]

    S8=32*( -255)/( -1)

    S8=32*255

    S8=8160

    2)
    I am not shure that this question have solution.

    Geometric progression in this case:

    Six terms:
    4,12,36,108,324,972

    4+12+36+108+324+972=1456

    Seven terms:
    4,12,36,108,324,972,2916

    4+12+36+108+324+972+2916=4372

  • math - ,

    In first question:

    Sn=S8=32*[(1-2^8)/(1-2)]

  • math - ,

    In a geometric progression,the product of the 2nd and 4th terms is double the 5th terms and the sum of the first four terms is 80.find the gp

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