# Thread: Sequences - Arithmetic and Recursively Defined

1. ## Sequences - Arithmetic and Recursively Defined

There are two problems that are bothering me on my homework. One is this:

The other is a story problem:

An architect is asked to design a theater with 20 seats in the first row, 24 in the second row, 28 in the third, and so on. If the theater is to have a capacity of 2480 seats, how many rows must there be?

Can anyone help with these two?

2. Originally Posted by blacksuzaku

An architect is asked to design a theater with 20 seats in the first row, 24 in the second row, 28 in the third, and so on. If the theater is to have a capacity of 2480 seats, how many rows must there be?
This looks like an arithmetic sequence - remember that in an arithmetic sequence, a common difference is added (or subtracted) each time. Here, we're adding 4 each time. To find the sum of the first n terms of an arithmetic sequence, we use:

$\displaystyle S_{n} = \frac{n(a_{1} + a_{n})}{2}$.

Also, $\displaystyle a_{n} = a_{1} + d(n-1)$ where d is the common difference.

Combining these, along with the fact that we know $\displaystyle a_{1} = 20$, the sum, $\displaystyle S_{n} = 2480$ and d, the common difference, is 4, gives:

$\displaystyle 2480 = \frac{n[20 + 20 + 4(n-1)]}{2}$.

Now, solve for n; you'll get a quadratic which factors, although the quadratic formula is probably the easier way to go.

3. First bit...

$\displaystyle a_2 = \frac{2(2^3) + 2}{3(2^2)} = \frac{3}{2}$

Then reinsert 3/2 to get the next one, etc...

4. I did that, Sean... but when I hit a6, I ended up with a number that had an insane amount of digits. ._.;;

5. Keep it in fraction form.

6. Hello, blacksuzaku!

Here's the second one . . .

An architect is asked to design a theater with 20 seats in the first row,
24 in the second row, 28 in the third, and so on. If the theater is to have
a capacity of 2480 seats, how many rows must there be?

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

. . where: .$\displaystyle a$ = first term, $\displaystyle d$ = common difference, $\displaystyle n$ = number of terms.

We have an arithmetic series: .$\displaystyle S_n \;=\;20 + 24 + 28 + 32 + \hdots\;=\;2480$

It has: .$\displaystyle a = 20,\;\;d = 4,\;\;n\text{ terms}$

So we have: .$\displaystyle \frac{n}{2}\bigg[2(20) + (n-1)4\bigg] \:=\:2480$

. . which simplifies to: .$\displaystyle n^2 + 9n - 1240 \:=\:0$

. . which factors: .$\displaystyle (n-31)(n+40) \:=\:0$

. . and has the positive root: .$\displaystyle n \:=\:31$

Therefore, the theater will have 31 rows.

7. Originally Posted by blacksuzaku
There are two problems that are bothering me on my homework. One is this:

The other is a story problem:

An architect is asked to design a theater with 20 seats in the first row, 24 in the second row, 28 in the third, and so on. If the theater is to have a capacity of 2480 seats, how many rows must there be?

Can anyone help with these two?
For the first one it is just plug and chug... for n=2 we get

$\displaystyle a_2=\frac{2a_1^3+2}{3a_1^2}=$

Since we $\displaystyle a_1=2$ we can just plug it in to get

$\displaystyle a_2=\frac{2(2)^3+2}{3(2)^2}=\frac{18}{12}=\frac{3} {2}$

Now that we know $\displaystyle a_2=\frac{3}{2}$ we can find the next term... just rinse and repeat...

For the 2nd we are dealing with an arithmetic series and we will need the formula

$\displaystyle S_n=\frac{n(2a_1+(n-1)d)}{2}$ where $\displaystyle a_1$ is the first term and d is the difference between sequence members

So for us we know that $\displaystyle S_n=2480 \\\ a_1=20 \\\ d=4$

So plugging all of this in we get

$\displaystyle 2480=\frac{n(2(20)+(n-1)4)}{2} \iff 4960=40n+4n^2-4n \iff 4n^2+36n-4960=0$

Now factoring a bit we get

$\displaystyle 4[n^2+9n-1240]=0 \iff 4(n+40)(n-31)$ so we get n=-40 and n=31.

Since we can't build a negative number of rows we get that we need 31 rows.

I hope this helps.

Good luck.

Edit: Too slow. sigh

8. Let's just say that the answer I got for a5 was 2935497269576521/2329904227757400

And I still have to plug that into a6... see my problem?

9. I suggest you use the Ans button on your calculator to refer to the previous term and give each term to 3 significant figures.

10. ## Spred sheet

I ran it in excel with 12 iterations and this is what I came up with

11. Well that merits the proverbial roflcopter... I feel stupid now :P

Stupid TI-89 and its "ZOMG lets keep everything in rational form"...