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geometric progession

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
Reply 1
Avatar for simon0
simon0
13
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?
Reply 2
Avatar for gchad88
gchad88
OP
2
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
Reply 3
Avatar for simon0
simon0
13
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.
Reply 4
Avatar for gchad88
gchad88
OP
2
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
Reply 5
Avatar for simon0
simon0
13
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.
Reply 6
Avatar for simon0
simon0
13
How did it go?
Reply 7
Avatar for gchad88
gchad88
OP
2
Struggling! unfortunately maths isn't my strong point!..as you may be able to tell:biggrin: . ill have another go tonight!

cheers
Reply 8
Avatar for gchad88
gchad88
OP
2
16000n+1400n^2-1400n=2Sn is the next part hopefully?
Reply 9
Avatar for simon0
simon0
13
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.
Reply 10
Avatar for gchad88
gchad88
OP
2
16000n+-1400n+1400n^2=7600000

14600n+1400n^2=7600000?
Reply 11
Avatar for simon0
simon0
13
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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