WHAT AM I DOING WRONG?

Maryanne

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Jan 11, 2006
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I have been trying this problem for awhile but i can't see to get it.

QA hockey arena has a total seating capacity of 15 690. The first row of seats around the rink has 262 seats. The number of seats in each subsequent rowincreases by 18.

a) How mnay rows of seats are in the arena?

b) By what percent would the capacity increase if three more rows were added after the last row?


a) This is what I did;

a= 262

tn= 15 690

d=18

n= ?


tn= a+(n-1)d


15 690 = 262+(n-1)(18)

15 690 =262+18n-18

15 690= 244+18n

15446= 18n

858.1111 =n

That doesn't make sense, you can't have .111 of a seat!
 
Maryanne said:
I have been trying this problem for awhile but i can't see to get it.

QA hockey arena has a total seating capacity of 15 690. The first row of seats around the rink has 262 seats. The number of seats in each subsequent rowincreases by 18.

a) How mnay rows of seats are in the arena?

b) By what percent would the capacity increase if three more rows were added after the last row?


a) This is what I did;

a= 262

tn= 15 690

d=18

n= ?


tn= a+(n-1)d


15 690 = 262+(n-1)(18)

15 690 =262+18n-18

15 690= 244+18n

15446= 18n

858.1111 =n

That doesn't make sense, you can't have .111 of a seat!

Your work looks very good. Since n represents the number of rows, not the number of seats, I see no reason there couldn't be 858 rows.
 
Great, thanks for the reassurance, but what about the second part of the question?

By what percent would the capacity increase if three more rows were added after the last row?
 
Hello, Maryanne!

A hockey arena has a total seating capacity of 15 690.
The first row of seats around the rink has 262 seats.
The number of seats in each subsequent row increases by 18.

a) How mnay rows of seats are in the arena?

b) By what percent would the capacity increase
if three more rows were added after the last row?


a) This is what I did;

\(\displaystyle a\,=\,262,\;\;\;t_n\,=\,15,690,\;\;\;d\,=\,18,\;\;\;n\,=\,?\)

\(\displaystyle t_n\:=\:a\,+\,(n\,-\,1)d\)

\(\displaystyle 15,690\:=\:262\,+\,(n\,-\,1)(18)\;\) . . . this is wrong
You said: there 15,690 seats in the n<sup>th</sup> row.

There are 15,690 seats in the entire arena.
The sum of the first n terms of an arithmetic series is: \(\displaystyle \,S_n\:=\:\frac{n}{2}\left[2a\,+\,d(n\,-\,1)]\\right]\)

So we have: \(\displaystyle \,\frac{n}{2}\left[2\cdot262\,+\,18(n\,-\,1)\right]\;=\;15,690\)

\(\displaystyle \;\;\)This simplifies to the quadratic: \(\displaystyle \,9n^2\,+\,253n\,-\,15,690\;=\;0\)

\(\displaystyle \;\;\)which factors: \(\displaystyle \.(n\,-\,36)(9n\,+\,523)\:=\:0\)

\(\displaystyle \;\;\)and has roots: \(\displaystyle \,n\,=\,36,\;n\,=\,-\frac{523}{9}\)

(a) Since we can't have negative or fractional rows, there are 36 rows in the area.


If three more rows are added, \(\displaystyle n\,=\,33\)
\(\displaystyle \;\;S_{33}\;=\;\frac{33}{2}\left[2\cdot262\,+\,32(18)\right]\;=\;18,150\)

There are: \(\displaystyle 18,150\,-\,15,690\:=\:2,460\) more seats.

The increase is: \(\displaystyle \,\frac{2,460}{15,690}\:=\:0.156787763\:\approx\:15.7\%\)
 
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