Results 1 to 12 of 12

Math Help - Geometric Series

  1. #1
    Junior Member
    Joined
    Feb 2007
    Posts
    33

    Geometric Series

    Determine whether or not the indicated geometric series converges. If so, find the value to which it converges.

    t1 = 81 and t5 = 1 (Two answers)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,803
    Thanks
    692
    Hello, Trentt!

    Determine if this geometric series converges.
    If so, find the value to which it converges.

    t
    1 = 81 and t5 = 1 (Two answers)

    The first term is: .t
    1 .= .a .= .81 .[1]

    The fifth term is: .t
    5 .= .ar^4 .= .1 .[2]

    . . . . . . . . . . . . . ar^4 . . . . 1
    Divide [2] by [1]: .------ .= . ---- . . . . r^4 .= .1/81
    . . . . . . . . . . . . . . a . . . . . 81

    . . Hence: .r .= .1/3

    In both cases, |r| < 1 . . . Therefore, both series converge.


    If r = 1/3, the sum is: .S .= .81/(1 - 1/3) .= .243/2

    If r = -1/3, the sum is: .S .= .81/(1 + 1/3) .= .243/4

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2007
    Posts
    33
    I have no idea what you just did. I don't understand how you get the ratio with just two terms
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Trentt View Post
    I have no idea what you just did. I don't understand how you get the ratio with just two terms
    You divide two sides,

    For example,

    x^2 = 7y^2
    x=y

    Divide left side with left side, divide right side with right side.And keep = sign in middle,

    Thus,

    Code:
    x^2      7y^2
    ----  =  -------
    x             y
    And you get,
    x=7y
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2007
    Posts
    33
    Nope, that didn't clear it up.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Trentt View Post
    Nope, that didn't clear it up.
    Here you do this one.

    xy^2 = xy^2
    And,
    x=y

    Divide both sides like I did and see what you get.
    Can you do this problem?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Feb 2007
    Posts
    33
    Nope.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Trentt View Post
    Nope.
    Did you try doing the problem. Post what you know how to do.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Feb 2007
    Posts
    33
    In the prior problems, before

    t1 = 81 and t5 = 1

    they gave us t1 and r, which I understood how to do. For this problem, I assume you have to find r, since the formula is t1/1-r
    I don't know how to find r.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Trentt View Post
    In the prior problems, before

    t1 = 81 and t5 = 1

    they gave us t1 and r, which I understood how to do. For this problem, I assume you have to find r, since the formula is t1/1-r
    I don't know how to find r.
    ok, so i don't see what's so hard to understand about Soroban's solution, but let's baby step through it.

    in a geometric series, we have all terms of the form ar^(n-1) where a is the first term, r is the common ratio and n = 1,2,3,4,5....

    so the first term is ar^(1-1) = ar^0 = a
    the second term is ar^(2-1) = ar^1 = ar
    the third term is ar^(3-1) = ar^2
    the fourth term is ar^(4-1) = ar^3
    the fifth term is ar^(5-1) = ar^4
    and so on...

    so from above, we can write the first term as a, so we have t1 = a, and we can write the fifth term as ar^4.

    so t1 = 81 and t5 = 1 so t5/t1 = 1/81, but also t1 = a and t5 = ar^4, so we have:

    .t5....=....1....=....ar^4
    ----......----.......------
    .t1.........81..........a

    so we get 1/81 = ar^4/a and we solve for r
    on the right, the a's cancel, so we are left with 1/81 = r^4
    so we get r = fourthroot(1/81) = +/- 1/3

    with me so far?
    Last edited by Jhevon; April 6th 2007 at 09:44 AM.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    anyway, now you have two values for r, so you can substitute them into that formula you mentioned to find the two answers. but Soroban already did that for you
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Feb 2007
    Posts
    33
    YES. Thank you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Taylor Series using Geometric Series.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 1st 2010, 04:17 PM
  2. Replies: 3
    Last Post: September 29th 2010, 06:11 AM
  3. Geometric Progression or Geometric Series
    Posted in the Math Topics Forum
    Replies: 8
    Last Post: October 8th 2009, 07:31 AM
  4. Replies: 11
    Last Post: April 1st 2008, 12:06 PM
  5. Replies: 2
    Last Post: January 23rd 2007, 08:47 AM

Search Tags


/mathhelpforum @mathhelpforum