If "n+5, 2n+1, and 4n-3" are three consequtive terms in an arithmetic sequence, find n.
Hello, shawli!
If $\displaystyle n+5,\; 2n+1,\; 4n-3$ are three consequtive terms
in an arithmetic sequence, find $\displaystyle n.$
$\displaystyle \text{We have: }\;\underbrace{n+5}_a,\:\underbrace{2n+1}_{a+d},\: \underbrace{4n-3}_{a+2d}$
Then: .$\displaystyle \begin{array}{cccccccccc}n+5 + d &=& 2n+1 && \Rightarrow && n - d &=& 4 & [1] \\
2n+1+d &=& 4n-3 && \Rightarrow && 2n-d &=& 4 & [2] \end{array}$
Subtract [2] - [1]: .$\displaystyle \boxed{n \:=\:0}$
Hi shawli,
Let's determine a common difference by subtracting the first term from the second term.
$\displaystyle 2n+1-(n+5)=2n+1-n-5=\boxed{n-4} $
Now determine the common difference between the 2nd and 3rd terms.
$\displaystyle 4n-3-(2n+1)=4n-3-2n-1=\boxed{2n-4} $
If it's truely an arithmetic sequence, these two common differences have to be equal.
$\displaystyle 2n-4=n-4$
Note to Anonymous1: I don't think 3 works.