# Thread: arithmetic sequence problem

1. ## arithmetic sequence problem

If "n+5, 2n+1, and 4n-3" are three consequtive terms in an arithmetic sequence, find n.

2. By inspection.

$\displaystyle n=3$

3. 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]: .$\displaystyle \boxed{n \:=\:0}$

4. Originally Posted by shawli
If "n+5, 2n+1, and 4n-3" are three consequtive terms in an arithmetic sequence, find n.
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.

5. Originally Posted by masters
Note to Anonymous1: I don't think 3 works.
Oh, I guess they have to be in the order given, then? Because $\displaystyle n=3$ yields the three consecutive integers $\displaystyle 7, 8, 9.$

6. Originally Posted by Anonymous1
Oh, I guess they have to be in the order given, then? Because $\displaystyle n=3$ yields the three consecutive integers $\displaystyle 7, 8, 9.$
That was my assumption. Your order would've been 8, 7, 9 which wouldn't work.

7. Originally Posted by Soroban
Hello, shawli!

$\displaystyle \text{We have: }\;\underbrace{n+5}_a,\:\underbrace{2n+1}_{a+d},\: \underbrace{4n-3}_{a+2d}$

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]: .$\displaystyle \boxed{n \:=\:0}$

How can n=0??????????????

8. Originally Posted by Anonymous1
How can n=0??????????????
If you do what Soroban said, n is obviously equal to 0.

You can review my post which is slightly different than Soroban's, but the conclusion is the same.

n = 0

Using that fact, the sequence becomes 5, 1, -3

9. Ah , the answer actually IS n=0, but I came to the same conclusion as Anonymous1 and thought it didn't make sense.

But now I get it ! Thanks!

### if three successive terms of a series in arithmetic progression are n 5,2n 3 and 4n-3

Click on a term to search for related topics.