# Help needed finding an equation.

• June 22nd 2010, 07:36 AM
rhawkin2
Help needed finding an equation.
I know this is probably elementary for the users of this site, but I have a problem that I am struggling to even find an equation for, let alone the solution. Here is the problem:

You have \$135,245,210. How many items can you purchase if they cost \$221,400 and their price increases by \$1800 with each item purchased?

It's been quite some time since I've done any Algebra, and I cannot even come up with the equation to find the solution. So any help would be much appreciated. Thanks!
• June 22nd 2010, 07:44 AM
wonderboy1953
You need an arithmetical formula to solve with.
• June 22nd 2010, 07:46 AM
rhawkin2
Any idea what that formula might be?
• June 22nd 2010, 07:54 AM
wonderboy1953
Check out this site: Mathwords: Arithmetic Sequence
• June 22nd 2010, 08:10 AM
rhawkin2
Thanks for the link. So, if I'm interpreting this correctly, the formula for this particular problem would read as follows:

135,245,210 = 221,400 + (n - 1)1800

Is that correct?
• June 22nd 2010, 08:16 AM
wonderboy1953
Quote:

Originally Posted by rhawkin2
Thanks for the link. So, if I'm interpreting this correctly, the formula for this particular problem would read as follows:

135,245,210 = 221,400 + (n - 1)1800

Is that correct?

Essentially correct. Keep in mind that you're increasing n step by step so you need another shortcut formula (a summation formula) which is at
Mathwords: Arithmetic Series
• June 22nd 2010, 08:20 AM
skeeter
Quote:

You have \$135,245,210. How many items can you purchase if they cost \$221,400 and their price increases by \$1800 with each item purchased?

sum of items purchased ...

$221400 + (221400+1800) + [221400+2(1800)] + [221400+3(1800)] + ... + [221400+n(1800)] = 135245210$

... see what you can do with this.
• June 22nd 2010, 08:59 AM
rhawkin2
Okay, now I'm really confused. Here's what I've got using Arithmetic Series:

an = 221,400 + (n-1)1800 = 219,600 + 1800n

Sum = n (221,400 + 219,600 + 1800n / 2)

Am I way off? And if not, how do I now solve for n?
• June 22nd 2010, 09:21 AM
blenyo
This is an AP
1st term, a = 221,400
Common difference, d = 1800
Last term, L = 135,245,210
But L = a + (n-1)d
=> 135,245,210 = 221,400 + (n-1)1800
221,400 + 1800n - 1800
219,600 + 1800n
n = (135,245,210 - 219,600)/1800 = 75
• June 22nd 2010, 10:04 AM
rhawkin2
That can't be right. Even assuming that formula is correct, (135,245,210 - 219,600)/1800 = 75014.22777, not 75.
• June 23rd 2010, 07:42 AM
blenyo
Agreed, that is a computational error, but you can use your calculator and get the answer.
• June 23rd 2010, 01:44 PM
Wilmer
Quote:

Originally Posted by rhawkin2
You have \$135,245,210. How many items can you purchase if they cost \$221,400 and their price increases by \$1800 with each item purchased?

OK, here we go; formula for sum of series is:
[number of terms] * [(first term + last term) / 2]

n = number of terms = ?
a = 1st term = 221400
f = last term = 221400 + 1800(n-1) : YES, n-1, not n; 1800 starts at n=2

n[(a + f)/2] = 135245210
n[((221400 + 221400+1800(n-1))/2 = 135245210
Solve for n: you'll get n = 284.0448.... ;

You can solve for Sum using n = 284 to get 135,212,400; a bit lesser than the 135,245,210.

As a general case:
s = sum
a = 1st term
b = constant increase
n = number of terms

n = [b - 2a + SQRT(4a^2 - 4ab + b^2 + 8bs)] / [2b]