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Math Help - Help needed finding an equation.

  1. #1
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    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!
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  2. #2
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    You need an arithmetical formula to solve with.
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  3. #3
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    Any idea what that formula might be?
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  4. #4
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    Check out this site: Mathwords: Arithmetic Sequence
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  5. #5
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    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?
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  6. #6
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    Quote Originally Posted by rhawkin2 View Post
    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
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  7. #7
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    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.
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  8. #8
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    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?
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  9. #9
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    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
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  10. #10
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    That can't be right. Even assuming that formula is correct, (135,245,210 - 219,600)/1800 = 75014.22777, not 75.
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  11. #11
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    Agreed, that is a computational error, but you can use your calculator and get the answer.
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  12. #12
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    Quote Originally Posted by rhawkin2 View Post
    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]
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