Results 1 to 3 of 3

Math Help - Geometric Progression- Ans to confirm only

  1. #1
    Member
    Joined
    Mar 2008
    Posts
    204

    Geometric Progression- Ans to confirm only

    Question:
    if 1+y, 1+3y, 1+4y (y cant be 0..sorry i dont know how to write it in math equation here ) are three consecutive terms of a Geometric Progression. Find
    (a) the value of y
    (b) the sum to infinity.

    the ans from book
    (a) -1/5
    (b) 8/5

    but i have calculate and get this ans
    (a) 1/5
    (b) -18/5

    thank for your advice
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,653
    Thanks
    1478
    Quote Originally Posted by nikk View Post
    Question:
    if 1+y, 1+3y, 1+4y (y cant be 0..sorry i dont know how to write it in math equation here ) are three consecutive terms of a Geometric Progression. Find
    (a) the value of y
    (b) the sum to infinity.

    the ans from book
    (a) -1/5
    (b) 8/5

    but i have calculate and get this ans
    (a) 1/5
    (b) -18/5

    thank for your advice
    In a geometric progression: \frac{t_{n + 1}}{t_n} = r.


    So \frac{1 + 3y}{1 + y} = r and \frac{1 + 4y}{1 + 3y} = r.


    Therefore \frac{1 + 3y}{1 + y} = \frac{1 + 4y}{1 + 3y}

    (1 + 3y)^2 = (1 + y)(1 + 4y)

    1 + 6y + 9y^2 = 1 + 5y + 4y^2

    5y^2 + y = 0

    y(5y + 1) = 0

    y = 0 or 5y + 1 = 0

    y = 0 or y = -\frac{1}{5}.


    Since y \neq 0 that means y = -\frac{1}{5}.


    Now to find the sum to infinity, we need a and r


    r = \frac{1 + 3y}{1 + y}

     = \frac{1 + 3\left(-\frac{1}{5}\right)}{1 - \frac{1}{5}}

     = \frac{1 - \frac{3}{5}}{\frac{4}{5}}

     = \frac{\frac{2}{5}}{\frac{4}{5}}

     = \frac{1}{2}.


    Are we assuming that a = 1 + y?

    If so a = \frac{4}{5}.


    Now, since |r| < 1 we can work out the sum to infinity...

    S_{\infty} = \frac{a}{1 - r}

     = \frac{\frac{4}{5}}{1 - \frac{1}{2}}

     = \frac{\frac{4}{5}}{\frac{1}{2}}

     = \frac{8}{5}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2008
    Posts
    204
    tq for the post ..now i can see where my calculation wrong

    can u guide me how to make this equation in latex look more bigger

    http://www.mathhelpforum.com/math-he...ze-bigger.html
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Geometric Progression
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: May 7th 2011, 11:44 PM
  2. Geometric Progression or Geometric Series
    Posted in the Math Topics Forum
    Replies: 8
    Last Post: October 8th 2009, 07:31 AM
  3. Geometric Progression
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 21st 2009, 07:50 PM
  4. Replies: 8
    Last Post: March 23rd 2009, 07:26 AM
  5. arithmetic progression--ans to confirm
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 14th 2009, 09:20 PM

Search Tags


/mathhelpforum @mathhelpforum