Results 1 to 4 of 4

Math Help - Geometric Sequence Problem

  1. #1
    Newbie
    Joined
    Jun 2011
    Posts
    12

    Geometric Sequence Problem

    "The sum of the first 2 terms of a geometric sequence is four times the sum of the next 2 terms, and the sum to infinity is 3. Determine 'r' and write down the sequences."

    I don't know how to start. Any pointers for what I need to do? Solutions welcome, this is prep for a test I'm writing in two hours. Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28

    Re: Geometric Sequence Problem

    Some equations to consider a+ar = 4(ar^2+ar^3) and \frac{a}{1-r}= 3
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,102
    Thanks
    68

    Re: Geometric Sequence Problem

    a + ar = 4(ar^2 + ar^3)
    a(1 + r) = 4ar^2(1 + r)
    a = 4ar^2
    1 = 4r^2
    r^2 = 1/4
    r = 1/2
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,719
    Thanks
    634

    Re: Geometric Sequence Problem

    Hello, Pulsys!

    The sum of the first 2 terms of a geometric sequence is four times the sum of the next 2 terms,
    and the sum to infinity is 3. .Determine r and write down the sequences.

    The first four terms are: . a,\,ar,\,ar^2,\,ar^3

    We are told: . \begin{Bmatrix}a + ar \:=\:4(ar^2 + ar^3)& [1] \\ \dfrac{a}{1-r} \:=\:3 & [2] \end{Bmatrix}\;\;\text{ Note that: }\,|r| < 1


    From [2], we have: . a \:=\:3(1-r)\;\;[3]


    From [1], we have: . a+ar \:=\:4ar^2 + 4ar^3 \quad\Rightarrow\quad 4ar^3 + 4ar^2 - ar - a \:=\:0

    . . . 4ar^2(r+1) - a(r+1) \:=\:0 \quad\Rightarrow\quad a(r+1)(4r^2 - 1) \:=\:0

    . . a(r+ 1)(2r-1)(2r+1) \:=\:0 \quad\Rightarrow\quad r \:=\:\rlap{//}\text{-}1,\:\tfrac{1}{2},\:\text{-}\tfrac{1}{2}


    Substitute into [3]: . \begin{Bmatrix}r = \tfrac{1}{2}\!: & a = \tfrac{9}{2} \\ \\[-3mm] r = \text{-}\tfrac{1}{2}\!: & a = \tfrac{3}{2} \end{Bmatrix}


    The sequences are: . \begin{Bmatrix} \frac{9}{2},\:\frac{9}{4},\:\frac{9}{8},\:\frac{9}  {16}\:\hdots \\ \\[-3mm] \frac{3}{2},\:\text{-}\frac{3}{4},\:\frac{3}{8},\:\text{-}\frac{3}{16}\:\hdots \end{Bmatrix}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. geometric and arithmetic sequence problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 5th 2010, 07:06 PM
  2. Geometric Sequence problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 5th 2010, 08:09 AM
  3. geometric sequence problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 25th 2008, 05:31 AM
  4. Geometric sequence problem
    Posted in the Algebra Forum
    Replies: 4
    Last Post: October 24th 2007, 10:58 PM
  5. Replies: 2
    Last Post: January 23rd 2007, 08:47 AM

Search Tags


/mathhelpforum @mathhelpforum