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    So many questions require a calculator; can literally not be done by a mere mortal such as myself. I'm talking about the ones where sum whatever is greater than/less than a certain number. In particular though:

    In the C1 revision guide there's this question,

    Given that the sum to n terms is greater than 1000, find the least possible value of n. (We know that 1st term, a, is 10, and common difference, d, is 1.5). The answer is 31, however if anyone could provide a workings toward this, please do so.

    Here's what I did

    n/2 [ 2a + (n-1)d] > 1000

    n [2a + (n-1)d] > 2000

    n [20 + 3/2(n-1)] > 2000

    n [40 + 3(n-1)] > 2000

    40n + 3n^2 - 3n > 2000

    3n^2 + 37n - 2000 > 0

    ^ how the heck you supposed to solve that by hand? Cannot even be reduced by simplification of 2 or 4 or something, ahhh.
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    (Original post by snow leopard)

    n [20 + 3/2(n-1)] > 2000

    n [40 + 3(n-1)] > 2000
    You made a little slip but I admit the final equation is a bit awkward.
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    (Original post by snow leopard)
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    it can't be factorised without the quadratic formula can it?

    anyway yeah i've seen a few questions with ridiculously large numbers in textbooks but I am yet to see a question like this on an actual C1 paper.

    edit: i got 31 but with the quadratic formula (with a calculator), they would not expect you to answer a question like this without a calculator in a C1 paper.
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    I'd say since you don't need the exact number (just the nearest whole), trial and error wouldn't be too hard for that tbh.
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    (Original post by fudgesundae)
    it can't be factorised without the quadratic formula can it?

    anyway yeah i've seen a few questions with ridiculously large numbers in textbooks but I am yet to see a question like this on an actual C1 paper.
    My teacher confirmed with the department a while ago that such questions wouldn't be featured in the actual exam and are ridiculous, but thought I might as well double check with the forum.
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    Well you can estimate it with a bit of ease:

    Since the answer is n<\frac{\sqrt{48289}}{\sqrt{36}  } - \frac{17}{6}

    And it'll only take a few minutes, but I guess it does depend on how accurate they want the answer.
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    (Original post by StephenP91)

    Since the answer is n<\frac{\sqrt{48289}}{\sqrt{36}  } - \frac{17}{6}
    It isn't you know.

    It's n>\frac{\sqrt{49369}-37}{6}.

    and it's not an easy calculation for the average C1 student.

    Edit:

    Thanks for the rep. It's cool. If you tell me why (and who you are) you can have the rep of your choice tomorrow. Don't all go saying it was you.
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