SOLUTION: There were 15 seats in the first row of the stadium. Each subsequent row has 2 additional seats per row. If there were 10 total rows, how many seats were in the stadium?

Algebra ->  Customizable Word Problem Solvers  -> Numbers -> SOLUTION: There were 15 seats in the first row of the stadium. Each subsequent row has 2 additional seats per row. If there were 10 total rows, how many seats were in the stadium?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1206662: There were 15 seats in the first row of the stadium. Each subsequent row has 2 additional seats per row. If there were 10 total rows, how many seats were in the stadium?
Found 2 solutions by Edwin McCravy, greenestamps:
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
If you knew nothing about arithmetic sequences, you would add this row
of 10 numbers:

15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33

But since you know about arithmetic sequences you can substitute in

S%5Bn%5D%22%22=%22%22expr%28n%2F2%29%282a%5B1%5D%2B%28n-1%29d%29

with a1 = 15, d = 2, and n = 10

S%5B10%5D%22%22=%22%22expr%2810%2F2%29%282%2815%29%5E%22%22%2B%2810-1%29%282%29%29
S%5B10%5D%22%22=%22%225%2830%2B%289%29%282%29%5E%22%22%29
S%5B10%5D%22%22=%22%225%2830%2B18%5E%22%22%29
S%5B10%5D%22%22=%22%225%2848%5E%22%22%29
S%5B10%5D%22%22=%22%22240%5E%22%22

The formula saved you from having to add 10 numbers together.

It wasn't that great a help here, but some problems will require
you to add 100 or even 1000 numbers. Then it will be a much bigger
help.  In school you are often given easier problems to show you
the principle, so you can use it on harder problems.

Edwin

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


The other tutor used the standard textbook form of the formula for finding the sum of the terms of an arithmetic sequence. While that formula is of course valid, I personally find it "ugly", because it is hard to see why the formula gives the right answer.

I find it far more satisfying to use a formula that is easy to understand.

From the definition of average, it follows that, for any set of numbers, the sum of the numbers is the number of terms, multiplied by the average of the terms.

For an arithmetic sequence, that way of finding the sum is easy, because we know the number of terms, and we know that in an arithmetic sequence the average of the terms is the average of the first and last terms. So we have

sum = (number of terms)*(average of first and last terms)

The number of terms in this problem is 10. The first term is 15 and the last (10th) term is 15 plus 9 times the common difference of 2, or 15+18=33.

So the sum of the terms is

%2810%29%28%2815%2B33%29%2F2%29=%2810%29%2848%2F2%29=10%2A24=240

ANSWER: 240