The numbers p,10 and q are 3 consecutive terms of an arithmetic progression .the numbers p,6 and q are 3 consecutive terms of a geometric progression .by first forming two equations in p and q show that p^2-20p+36=0
Hence find the values of p and q for which the geometric progression converges
Help please

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  1. from the first:
    10-p = q-10
    q = 20-p

    from the second:
    pq = 36

    by substitution,
    p(20-p) = 36
    20p - p^2 = 36
    p^2 - 20p + 36 = 0 ----> thus shown

    (p - 18)(p - 2) = 0
    p = 18 or p = 2
    then q = 36/18 = 2 or p = 36/2 = 18

    for p = 18, q = 2
    the AS is 18, 10 and 2
    the GS is 18, 6, 2 -----> which converges, r = 1/2

    for p = 2, q = 18
    the AS is 2, 10, 18
    the GS is 2, 6, 18, ---> diverges , r > 1

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