The 1st 3 terms of an A.P., a, a+d, and a+2d, are the same as the 1st 3 terms of a G.P.( a is not equal to 0). Show that this is only possib

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The 1st 3 terms of an A.P., a, a+d, and a+2d, are the same as the 1st 3 terms of a G.P.( a is not equal to 0). Show that this is only possible if r=1 and d=0

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Thái Dương 4 years 2021-08-26T19:16:12+00:00 1 Answers 4 views 0

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  1. Adela
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    2021-08-26T19:17:41+00:00
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    Answer:

    See Below.

    Step-by-step explanation:

    The first three terms of an A.P is equivalent to the first three terms of a G.P.

    We want to show that this is only possible if r = 1 and d = 0.

    If a is the initial term and d is the common difference, the A.P. will be:

    a, a+d, \text{ and } a+2d

    Likewise, for the G.P., if a is the initial term (and it does not equal 0) and r is the common ratio, then our sequence is:

    a, ar,\text{ and } ar^2

    The second and third terms must be equivalent. Thus:

    a+d=ar\text{ and } a+2d=ar^2

    We can cancel the d. Multiply the first equation by -2:

    -2a-2d=-2ar

    We can now add this to the second equation:

    (a+2d)+(-2a-2d)=(ar^2)+(-2ar)

    Simplify:

    -a=ar^2-2ar

    Now, we can divide both sides by a (we can do this since a is not 0):

    -1=r^2-2r

    So:

    r^2-2r+1=0

    Factor:

    (r-1)^2=0

    Thus:

    r=1

    The first equation tells us that:

    a+d=ar

    Therefore:

    a+d=a(1)\Rightarrow a+d=a

    Hence:

    d=0

    Q.E.D.

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