# Thread: Help needed finding an equation.

1. ## 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! 2. You need an arithmetical formula to solve with. 3. Any idea what that formula might be? 4. Check out this site: Mathwords: Arithmetic Sequence 5. 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? 6. 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 7. 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.

8. 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?

9. 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

10. That can't be right. Even assuming that formula is correct, (135,245,210 - 219,600)/1800 = 75014.22777, not 75.

11. Agreed, that is a computational error, but you can use your calculator and get the answer.

12. 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]