# Thread: Arithmetic Series

1. ## Arithmetic Series

So there is wall...the top has 1 brick, then the next row has 3 bricks, then 5 bricks etc. there are 2500 bricks in the wall then how many rows are there

im just really confused with arithmetic series right now and need general help with them too....

pref. without using some formula but a more explanative version

2. Originally Posted by stones44
So there is wall...the top has 1 brick, then the next row has 3 bricks, then 5 bricks etc. there are 2500 bricks in the wall then how many rows are there

im just really confused with arithmetic series right now and need general help with them too....

pref. without using some formula but a more explanative version

This would be a series of the odd numbers so it would be $\displaystyle \sum_{i=0}^{n}2i+1$

can you find n from that?

3. no thats what im saying..im very confused

4. Originally Posted by Mathstud29
This would be a series of the odd numbers so it would be $\displaystyle \sum_{i=0}^{n}2i+1$

can you find n from that?
$\displaystyle \sum_{n=0}^{x}2n+1=\sum_{n=0}^{x}2n+\sum_{n=0}^{x} 1=\frac{2x(x+1)}{2}+x$

Now solve $\displaystyle \frac{2x(x+1)}{2}+x=2500$

5. Hi
Originally Posted by Mathstud28
$\displaystyle \sum_{n=0}^{x}2n+1=\sum_{n=0}^{x}2n+\sum_{n=0}^{x} 1=\frac{2x(x+1)}{2}+x{\color{red}+1}$
There are $\displaystyle n-p+1$ elements in the set of positive integers $\displaystyle \{p,\,p+1,\,p+2,\,\ldots,\,n-1,\,n\}$, assuming $\displaystyle n\geq p$.

6. Hello, stones44!

So there is wall.
The top has 1 brick, then the next row has 3 bricks, then 5 bricks, etc.
There are 2500 bricks in the wall.
How many rows are there?

Without a formula? . . . Then a child can do it!

Just start adding: .$\displaystyle 1 + 3 + 5 + 7 + 9 + \hdots$ . until you reach 2500.

Then count up the number of numbers you added.

. . . I'll wait in the car . . .

Formula for the sum of the first $\displaystyle n$ terms of an arithmetic series:
. . $\displaystyle S_n \;=\;\frac{n }{2}[2a + (n-1)d]$

We have: .first term, $\displaystyle a = 1$ . . . common difference, $\displaystyle d = 2$

We know: .$\displaystyle S_n = 2500$ . . . and we want $\displaystyle n$.

So we have: .$\displaystyle 2500 \:=\:\frac{n}{2}[2(1) + (n-1)2] \quad\Rightarrow\quad 2500 \:=\:\frac{n}{2}(2n)\quad\Rightarrow\quad n^2 \:=\:2500$

Therefore: .$\displaystyle n \:=\:50$ . . . There are 50 rows.