Chenice
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I am trying to do question 10) iv) in this paper

http://www.ocr.org.uk/Images/63180-q...hematics-2.pdf

It involves finding the sum of the GP for cycling
I have r=1.1 a=2 n=30
In the formula book I have the sum of a GP as
Sn= a(1-r^n)/(1-r)
So plugging in I have got = 2(1-1.1^30)/(1.1.1)

However in the markscheme it says
2(1.1^30-1)/(1.1-1)

Why is this? :/
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Jai Sandhu
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(Original post by Chenice)
I am trying to do question 10) iv) in this paper

http://www.ocr.org.uk/Images/63180-q...hematics-2.pdf

It involves finding the sum of the GP for cycling
I have r=1.1 a=2 n=30
In the formula book I have the sum of a GP as
Sn= a(1-r^n)/(1-r)
So plugging in I have got = 2(1-1.1^30)/(1.1.1)

However in the markscheme it says
2(1.1^30-1)/(1.1-1)

Why is this? :/
2(1-1.1^30)/(1-1.1) is the same as 2(1.1^30-1)/(1.1-1)

All that has been done is you have multiplied top and bottom of the equation by -1

In essence the formulae can either be written:

Sn= a(1-r^n)/(1-r)

or

Sn= a(r^n-1)/(r-1)
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Chenice
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(Original post by Jai Sandhu)
2(1-1.1^30)/(1-1.1) is the same as 2(1.1^30-1)/(1.1-1)

All that has been done is you have multiplied top and bottom of the equation by -1

In essence the formulae can either be written:

Sn= a(1-r^n)/(1-r)

or

Sn= a(r^n-1)/(r-1)
Oh ok thank you!
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