Results 1 to 3 of 3

Math Help - Questions

  1. #1
    Junior Member
    Joined
    Jul 2008
    Posts
    26

    Questions

    if a^2,b^2,c^2 are in A.P. then 1/b+c,1/c+a,1/a+b ARE In??
    Will it be AP,GP,HP or none..??
    Last edited by cnpranav; April 2nd 2009 at 04:13 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    779
    Hello, cnpranav!

    If a^2,b^2,c^2 are in A.P. then: . \frac{1}{b+c},\;\frac{1}{c+a},\;\frac{1}{a+b} .are in ...

    Will it be AP, GP, HP or none?
    Since a^2,b^2,c^2 are in Arithmetic Progression,

    we have: . \begin{array}{ccccccccc}<br />
b^2-a^2 \:=\: d &\Rightarrow & (b-a)(b+a) \:=\:d & \Rightarrow & a+b \:=\:\frac{d}{b-a} \\<br />
c^2-b^2 \:=\:d & \Rightarrow & (c-b)(c+b) \:=\:d & \Rightarrow & b+c \:=\:\frac{d}{c-b} \end{array}
    . . . . . . . \begin{array}{ccccccccc}c^2-a^2 \:=\:2d & \Rightarrow & (c-a)(c+a) \:=\:2d & \Rightarrow & c+a \:=\:\frac{2d}{c-a} \end{array}


    \text{Then: }\;\left\{\frac{1}{b+c},\;\frac{1}{c+a},\;\frac{1}  {a+b}\right\}\;\text{ becomes: }\;\underbrace{\frac{c-b}{d}}_{u_1},\;\underbrace{\frac{c-a}{2d}}_{u_2},\;\underbrace{\frac{b-a}{d}}_{u_3}


    We find that: . \begin{array}{ccccc}u_2-u_1 &=&\dfrac{c-a}{2d} - \dfrac{c-b}{d} &=& \dfrac{2b-a-c}{2d} \\ \\[-4mm]<br />
u_3-u_2 &=& \dfrac{b-a}{d} - \dfrac{c-a}{2d} &=& \dfrac{2b-a-c}{2d} \end{array}


    u_1,u_2,u_3\,\text{ have a }common\;di\!f\!\!ference\;\hdots\;<br />
\text{ They are in }{\color{blue}A.P.}

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jul 2008
    Posts
    26

    Smile

    Quote Originally Posted by Soroban View Post
    Hello, cnpranav!

    Since a^2,b^2,c^2 are in Arithmetic Progression,

    we have: . \begin{array}{ccccccccc}<br />
b^2-a^2 \:=\: d &\Rightarrow & (b-a)(b+a) \:=\:d & \Rightarrow & a+b \:=\:\frac{d}{b-a} \\<br />
c^2-b^2 \:=\:d & \Rightarrow & (c-b)(c+b) \:=\:d & \Rightarrow & b+c \:=\:\frac{d}{c-b} \end{array}
    . . . . . . . \begin{array}{ccccccccc}c^2-a^2 \:=\:2d & \Rightarrow & (c-a)(c+a) \:=\:2d & \Rightarrow & c+a \:=\:\frac{2d}{c-a} \end{array}


    \text{Then: }\;\left\{\frac{1}{b+c},\;\frac{1}{c+a},\;\frac{1}  {a+b}\right\}\;\text{ becomes: }\;\underbrace{\frac{c-b}{d}}_{u_1},\;\underbrace{\frac{c-a}{2d}}_{u_2},\;\underbrace{\frac{b-a}{d}}_{u_3}


    We find that: . \begin{array}{ccccc}u_2-u_1 &=&\dfrac{c-a}{2d} - \dfrac{c-b}{d} &=& \dfrac{2b-a-c}{2d} \\ \\[-4mm]<br />
u_3-u_2 &=& \dfrac{b-a}{d} - \dfrac{c-a}{2d} &=& \dfrac{2b-a-c}{2d} \end{array}


    u_1,u_2,u_3\,\text{ have a }common\;di\!f\!\!ference\;\hdots\;<br />
\text{ They are in }{\color{blue}A.P.}

    thanks a ton
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. More log questions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 31st 2010, 05:58 PM
  2. Please someone help me with just 2 questions?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 4th 2009, 05:55 AM
  3. Some Questions !
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 3rd 2009, 04:09 AM
  4. Replies: 4
    Last Post: July 19th 2008, 08:18 PM
  5. Replies: 3
    Last Post: August 1st 2005, 02:53 AM

Search Tags


/mathhelpforum @mathhelpforum