Another arithmetic sequence problem

Hello, shawli!

You typed part (b) incorrectly . . .

Quote:

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

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

Answers: . $(a)\;\;\text{-}1,\quad (b)\;\; 0$

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

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

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

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

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

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

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

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

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