Word Problem: What is total number of bricks in wall?

[ryana]

The number of bricks in the bottom row of a brick wall is 49. The next row up from the bottom contains 47 bricks, and each subsequent row contains 2 fewer bricks than the row immediately below it. The number of bricks in the top row is 3. If the wall is one brick thick, what is the total number of bricks in the wall?

I understand 47 + 45 + 43 + ... + 3 = TOTAL but there should be a faster way to calculate this using algebra. How do you go about building that equation?

Question is taken from the CLEP College Algebra practice test.

 

[stapel]

I understand 47 + 45 + 43 + ... + 3 = TOTAL but there should be a faster way to calculate this...

47 + 3 = 50
45 + 5 = 50
43 + 7 = 50
....

:wink:

Eliz.

 

[ryana]

I'm not seeing it. Please explain more.

 

[skeeter]

Elizabeth added the two terms from both ends of the series.

hopefully, you're not color blind ...do you "see" it now?

47 + 45 + 43 + 41 + ... + 9 + 7 + 5 + 3

47 + 3 = 50
45 + 5 = 50
43 + 7 = 50
41 + 9 = 50

 

[Denis]

Formula for sum of 1st n odd numbers is n^2 ; and n = (last number + 1) / 2

So (49+ 1)/2 = 25 = n ; n^2 = ?
Don't forget to deduct 1 (not part of series; starts at 3)

Edit: 47 changed to 49 !!

 

[soroban]

Hello, ryana!

This is what Stapel was telling you . . .

The number of bricks in the bottom row of a brick wall is 49.
The next row up from the bottom contains 47 bricks,
and each subsequent row contains 2 fewer bricks than the row immediately below it.
The number of bricks in the top row is 3.
If the wall is one brick thick, what is the total number of bricks in the wall?

\(\displaystyle \begin{array}{cccc}\text{We want:}& S & = & \;3 \:+\: 5 \:+\: 7 \:+\: 9 + \hdots + 49 \\ \text{Then:} & S & = & 49 + 47 + 45 + 43 + \hdots + 3 \\ \\ \text{Add:} & 2S & = & \underbrace{52 + 52 + 52 + 52 + \hdots + 52}_{\text{24 terms}} \end{array}\)

\(\displaystyle \text{Therefore: }\:2S \:=\24)(52) \:=\:1248\quad\Rightarrow\quad \boxed{S \:=\:624}\)

 

[galactus]

An observation to make is that the sum of the first n odd integers is given by n^2.

\(\displaystyle 1+3+5+7+9+..........+(2n-1)=n\cdot\frac{1+(2n-1)}{2}=n^{2}\)

Since n=25, we just have \(\displaystyle 25^{2}-1=624\)


The above is just a general case of what Soroban and Stapel posted. It is the same way that Gauss supposedly, as a young lad, showed the sum of the first n integers was \(\displaystyle \frac{n(n+1)}{2}\)

 

[ryana]

That's an interesting way to calculate the results. Thanks

Algebra is pretty frustrating but there are some really neat things it can do.