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    The first term of an arithmetic series is 14
    Common difference = -6, and so is the sum of the first n terms.

    Find n.


    Could someone please explain how to calculate this? I tried to get two equations so that I could solve them simultaneously, but failed.

    I know this a really stupid question...but it's stumped me.
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    What is an expression for the sum of the first n terms?
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    Sn = n/2 [2a+(n-1)d]
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    Sn = n/2 [2a+(n-1)d]

    A = 14, D = -6.

    Sub it in.
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    Question:
    what does N stand for?
    l-last
    d-difference
    so whats n?
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    (Original post by v1oXx-)
    Sn = n/2 [2a+(n-1)d]

    A = 14, D = -6.

    Sub it in.

    Unless I miscalculated that would gives me 0=30n-6n^2-12, which doesn't give me the answer.
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    Course it does.

    Factorize.
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    (Original post by v1oXx-)
    Course it does.

    Factorize.
    It won't factorise;
    6(-2 + 5n + -1n^2) is what it looks like factorised.
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    Try 2 brackets?
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    (Original post by v1oXx-)
    Try 2 brackets?
    Because you can't.

    Anything else that I could do to solve this?
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    ive just skimmed this so you might have done it already and rearranged but shouldnt you be setting it equal to -6 rather than 0?

    as d = Sn = -6
 
 
 
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