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18 May, 18:54

Find four numbers that form a geometric progression such that the third term is greater than the first by 12 and the fourth is greater than the second by 36.

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  1. 18 May, 19:02
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    5, 4.5, 13.5 and 40.5

    Step-by-step explanation:

    Since the numbers are in geometric progression, their form is essentially:

    a, ar, ar^2 and ar^3

    Where a and r are first term and common ratio respectively.

    From the information given in the catalog:

    Third term is greater than the first by 12 while fourth is greater than second by 36.

    Let's now translate this to mathematics.

    ar^2 - a = 12

    ar^3 - ar = 36

    From 1, a (r^2 - 1) = 12 and 2:

    ar (r^2 - 1) = 36

    From 2 again r[a (r^2 - 1] = 36

    The expression inside square bracket looks exactly like equation 1 and equals 12.

    Hence, 12r = 36 and r = 3

    Substituting this in equation 1,

    a (9 - 1) = 12

    8a = 12

    a = 12/8 = 1.5

    Thus, the numbers are 1.5, (1.5 * 3), (1.5 * 9), (1.5 * 27) = 1.5, 4.5, 13.5 and 40.5
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