Another arithmetic sequence problem

• Apr 3rd 2010, 02:04 PM
shawli
Another arithmetic sequence problem
For a given arithmetic sequence, $\displaystyle u_n=m$ and $\displaystyle u_m=n$. Find

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

Could someone explain to me the method?
• Apr 3rd 2010, 04:38 PM
satx
I would think that Un + m would equal 2m since Un = m. Could you explain why that is not the case?
• Apr 3rd 2010, 04:45 PM
Soroban
Hello, shawli!

You typed part (b) incorrectly . . .

Quote:

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: .$\displaystyle u_{m+n} \;=\;n + n(\text{-}1) \;=\;0$