Sequences - Arithmetic and Recursively Defined

**blacksuzaku** · June 12th 2008, 03:47 PM

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?

**Mathnasium** · June 12th 2008, 03:58 PM

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:

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

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

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

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

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

**sean.1986** · June 12th 2008, 03:59 PM

First bit...

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

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

**blacksuzaku** · June 12th 2008, 04:06 PM

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

**sean.1986** · June 12th 2008, 04:10 PM

Keep it in fraction form.

**Soroban** · June 12th 2008, 04:10 PM

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

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

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

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

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

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

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

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

Therefore, the theater will have 31 rows.

**TheEmptySet** · June 12th 2008, 04:12 PM

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

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

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

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

Now that we know $a_2=\frac{3}{2}$ 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

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

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

So plugging all of this in we get

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

Now factoring a bit we get

$4[n^2+9n-1240]=0 \iff 4(n+40)(n-31)$ 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

**blacksuzaku** · June 12th 2008, 04:16 PM

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?

**sean.1986** · June 12th 2008, 04:20 PM

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

**TheEmptySet** · June 12th 2008, 04:23 PM

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

Sequences - Arithmetic and Recursively Defined-capture1.jpg

**blacksuzaku** · June 12th 2008, 04:28 PM

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

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

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