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    hi there,

    really struggling with this one can anyone help?:

    A company involved in shale gas exploration needs to provide cost estimates to their client with regard to the initial drilling operation. they charge a one-off set up fee of £200,000. the daily rate starts at £8000 and increases by £1400 each day. a total budget of £4m is assigned to the project.

    1/find the maximum number of days available for drilling

    i know i need to use: Sn = n/2(2a+(n-1)d) but am struggling solve.

    any help would be massively appreciated. thank you
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    For the title of this thread, I think you mean the "arithmetic series" as the difference between one day's cost to the next is an addition of a set value (rather the multiplied in the case of the geometric sequence).

    First, how much can we spend?
    We are told we have budget of £4 million and we have a one-off cost. Can we find this?
    Let us call this "Sn".

    Second, can we see what is happening from day one?
    Day one: 8,000,
    Day two: 8,000 + 1,400,
    Day three: (8,000 + 1,400) + 1,400 = 8,000 + 2*(1,400),
    .
    .
    .

    Now we need: (Cost of day one) + (Cost of day two) + (Cost of day three) +...  \leq Sn .



    Can you see a theme, use the formula, carry on from here?
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    Thanks Simon. The way I was working it through was a=8000, d=1400 and sn= 4000000. is that the right way to go?

    thanks
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    (Original post by gchad88)
    Thanks Simon. The way I was working it through was a=8000, d=1400 and sn= 4000000. is that the right way to go?

    thanks
    For the "Sn", note you also have a one off cost as well.

    The rest looks fine.
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    (Original post by simon0)
    For the "Sn", note you also have a one off cost as well.

    The rest looks fine.
    Sn=3800000=n/2(2*8000)+(n-1)*1400

    380000=n/2(16000)+(n-1)1400

    n(16000 + 1400(n-1))/2= SN

    n(16000+1400n-1400 = 2sn

    hopefully on the right track? but this is where I get stuck unfortunately
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    (Original post by gchad88)
    Sn=3800000=n/2(2*8000)+(n-1)*1400

    380000=n/2(16000)+(n-1)1400

    n(16000 + 1400(n-1))/2= SN

    n(16000+1400n-1400 = 2sn

    hopefully on the right track? but this is where I get stuck unfortunately
    Looks okay but be careful with your brackets.

    Multiply out, simplify and put every term to one side and you should see a familiar type of equation.
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    How did it go?
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    Struggling! unfortunately maths isn't my strong point!..as you may be able to tell . ill have another go tonight!

    cheers
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    16000n+1400n^2-1400n=2Sn is the next part hopefully?
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    (Original post by gchad88)
    16000n+1400n^2-1400n=2Sn is the next part hopefully?
    Looks good. Just substitute in Sn, place all terms to one side and (as you can tell), you have a quadratic!

    Then you simply solve for n.
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    16000n+-1400n+1400n^2=7600000

    14600n+1400n^2=7600000?
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    (Original post by gchad88)
    16000n+-1400n+1400n^2=7600000

    14600n+1400n^2=7600000?
    Looks good, now solve for n.

    Are you okay from here?
 
 
 

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