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Math Help - Progressions

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    Progressions

    The sum of the first 3 terms of a geometric progression is 13 times its first term.Find the possible values of the common ratio of the geometric progression.
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    Quote Originally Posted by mastermin346 View Post
    The sum of the first 3 terms of a geometric progression is 13 times its first term.Find the possible values of the common ratio of the geometric progression.
    HI

    S_3=13a , where a is the first term

    \frac{a(1-r^3)}{1-r}=13a

    Don worry bout the a because they will eventually cancel off.

    Solve for r then .
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  3. #3
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    Quote Originally Posted by mathaddict View Post
    HI

    S_3=13a , where a is the first term

    \frac{a(1-r^3)}{1-r}=13a

    Don worry bout the a because they will eventually cancel off.

    Solve for r then .
    You should also be able to cancel a immediately and then use the difference of two cubes to cancel (1-r) given that  r \neq 1

    (a-b)^3 = (a-b)(a^2+ab+b^2)
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  4. #4
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    Hello, mastermin346!

    The sum of the first 3 terms of a geometric progression is 13 times its first term.
    Find the possible values of the common ratio of the geometric progression.

    The first three terms are: . a,\:ar,\:ar^2

    Their sum is 13 times the first term: . a + ar + ar^2 \:=\:13a

    We have: . r^2 + r - 12 \:=\:0 \quad\Rightarrow\quad (r - 3)(r + 4) \:=\:0

    Therefore: . r \;=\;3\text{ or }-4

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