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    • Thread Starter
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    I'm struggling on a question
    that says
    in a geometric progression the sum of the first 2 terms is 108 and the sum of the third and fourth term is 12.
    find the two possible values of the common ratio and their corresponding first terms

    its worth 6 marks

    i have tried doing it myself but i cant figure out what formula to use
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    The first term in a g.p. is a

    The second term is ar

    The third term is ar^2 and so on.

    Does that help?
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    (Original post by Mr M)
    The first term in a g.p. is a

    The second term is ar

    The third term is ar^2 and so on.

    Does that help?
    i know that already..what do i do with that?
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    (Original post by princess271)
    i know that already..what do i do with that?
    a + ar = 108

    ar2 + ar3 = 12

    Those are your simultaneous equations. Find 'a' and 'r'.
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    (Original post by thegodofgod)
    a + ar = 108

    ar2 + ar3 = 12

    Those are your simultaneous equations. Find 'a' and 'r'.
    thanks
    i actually wrote down exactly what you have given me then didnt know what to do
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    (Original post by princess271)
    thanks
    i actually wrote down exactly what you have given me then didnt know what to do
    nor do I lol

    EDIT:

    a + ar = 108

    ar2 + ar3 = 12

    a --> ar2 = multiplying by r2 (same with other).

    Hence:

    12 = 108 r2

    r2 = 12/108

    r2 = 1/9

    r = +- 1/3

    Then sub that into any of the equations to get the 'a' value.
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    (Original post by princess271)
    thanks
    i actually wrote down exactly what you have given me then didnt know what to do
    do i times a + ar = 108 by three
    • PS Helper
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    Just to expand on what thegodofgod said, notice you can take a factor of a out in the first case or ar² in the second case. Then you can divide one equation from the other to let you find r, and then substitute back to find a.

    And on another note, this isn't geometry (although it is a geometric progression... but for some reason geometry and geometric sequences/series don't have a great deal to do with each other).

    EDIT: Did I take ages to respond or were the replies before me really quick?
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    (Original post by thegodofgod)
    nor do I lol
    lmao
    u dont know what to do?
    i'll just ask my teacher tomorow then
    hes gona go mad at me though
    i hand in all my assignments late
    thanks for trying to help me
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    (Original post by nuodai)
    Just to expand on what thegodofgod said, notice you can take a factor of a out in the first case or ar² in the second case. Then you can divide one equation from the other to let you find r, and then substitute back to find a.

    And on another note, this isn't geometry (although it is a geometric progression... but for some reason geometry and geometric sequences/series don't have a great deal to do with each other).
    thanks
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    (Original post by nuodai)
    Just to expand on what thegodofgod said, notice you can take a factor of a out in the first case or ar² in the second case. Then you can divide one equation from the other to let you find r, and then substitute back to find a.

    And on another note, this isn't geometry (although it is a geometric progression... but for some reason geometry and geometric sequences/series don't have a great deal to do with each other).
    i still dont understand
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    (Original post by princess271)
    do i times a + ar = 108 by three
    no.

    you have:
    1) a + ar = 108
    2) ar2 + ar3 = 12

    Divide 2 by 1 to get
    \dfrac{1 + r}{r^2 + r^3} = \dfrac{1}{9}
    Solve for r.

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    (Original post by thegodofgod)
    nor do I lol

    EDIT:

    a + ar = 108

    ar2 + ar3 = 12

    a --> ar2 = multiplying by r2 (same with other).

    Hence:

    12 = 108 r2

    r2 = 12/108

    r2 = 1/9

    r = +- 1/3

    Then sub that into any of the equations to get the 'a' value.
    is a +- 162
    • Thread Starter
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    (Original post by nuodai)
    Just to expand on what thegodofgod said, notice you can take a factor of a out in the first case or ar² in the second case. Then you can divide one equation from the other to let you find r, and then substitute back to find a.

    And on another note, this isn't geometry (although it is a geometric progression... but for some reason geometry and geometric sequences/series don't have a great deal to do with each other).

    EDIT: Did I take ages to respond or were the replies before me really quick?
    the replies before you were quick
    • Thread Starter
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    is a +- 162?
    • Thread Starter
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    everyones disappeared now
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    (Original post by princess271)
    is a +- 162
    No: If you take the positive 1/3,

    a + ar = 108

    a(1+r) = 108

    a = 108 / (1+r)

    a = 108 / (1 + 1/3)

    a = 108 / (4/3)

    a = 108 x (3/4)

    a = 81
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    (Original post by thegodofgod)
    No: If you take the positive 1/3,

    a + ar = 108

    a(1+r) = 108

    a = 108 / (1+r)

    a = 108 / (1 + 1/3)

    a = 108 / (4/3)

    a = 108 x (3/4)

    a = 81
    i just realised your the TSR people..:/
    if i realised wouldnt have asked you so many question lol
    normally you people give me warnings
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    (Original post by princess271)
    i just realised your the TSR people..:/
    if i realised wouldnt have asked you so many question lol
    normally you people give me warnings
    I'm not one of the TSR people lol - I've only made loads of posts (1500+) so I've automatically got a really cool title - 'Peer Of The TSR Realm'
    • PS Helper
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    (Original post by princess271)
    i just realised your the TSR people..:/
    if i realised wouldnt have asked you so many question lol
    normally you people give me warnings
    We don't give you warnings because we're not moderators :p:
 
 
 
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