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Thread: Another arithmetic sequence problem

  1. #1
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    Another arithmetic sequence problem

    For a given arithmetic sequence, $\displaystyle u_n=m$ and $\displaystyle u_m=n$. Find

    a) the common difference.
    b) $\displaystyle u_n+m$.


    Answers: a) -1, b) 0

    Could someone explain to me the method?
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  2. #2
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    I would think that Un + m would equal 2m since Un = m. Could you explain why that is not the case?
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  3. #3
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    Hello, shawli!

    You typed part (b) incorrectly . . .


    For a given arithmetic sequence: . $\displaystyle \begin{array}{ccc}u_n&=&m \\ u_m &=& n \end{array}$

    Find: .$\displaystyle \text{(a) the common difference }$
    . . . . .$\displaystyle (b)\;u_{m+n}$

    Answers: .$\displaystyle (a)\;\;\text{-}1,\quad (b)\;\; 0 $
    The $\displaystyle k^{th}$ term is given by: .$\displaystyle u_k \;=\; a + (k-1)d$ . where: .$\displaystyle \begin{Bmatrix}a &=& \text{first term} \\ d &=& \text{common diff.} \end{Bmatrix}$


    We have: .$\displaystyle \begin{array}{ccccccccc}u_n \:=\:m & \;\Rightarrow\; & a+(n-1)d \:=\:m & \;\Rightarrow\; & a \:=\:m-(n-1)d & [1] \\ u_m \:=\:n & \Rightarrow & a+(m-1)d \:=\:n & \Rightarrow & a \:=\:n-(m-1)d & [2] \end{array}$


    Equate [1] and [2]: .$\displaystyle m - (n-1)d \:=\:n-(m-1)d \quad\Rightarrow\quad m-dn \:=\:n - dm $

    . . . . . . $\displaystyle m - n + dm - dn \:=\:0 \quad\Rightarrow\quad (m-n) + d(m-n) \:=\:0 \quad\Rightarrow\quad (m-n)(d+1) \:=\:0$


    If $\displaystyle m-n\:=\:0 \quad\Rightarrow\quad m \:=\:n $, we have: .$\displaystyle u_m \:=\:m$
    . . and the arithmetic sequence cannot be determined.

    If $\displaystyle d+1 \:=\:0$, then: .$\displaystyle \boxed{d \:=\:-1}$



    $\displaystyle u_{m+n} \;\;=\;\;a + (m+n-1)d \;\;=\;\;a + md + nd - d $

    . . . .$\displaystyle =\;\;a + md - d + nd \;\;=\;\;\underbrace{a + (m-1)d}_{\text{This is }u_m \,=\, n} + nd \;\;=\;\;n + nd$


    Since $\displaystyle d = \text{-}1$, we have: .$\displaystyle u_{m+n} \;=\;n + n(\text{-}1) \;=\;0$

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