6. A car was purchased for £18 000 on 1st January. On 1st January each following year, the value of the car is 80% of its value on 1st Januaryin the previous year.
(a) Show that the value of the car exactly 3 years after it was purchased is £9216.(1)The value of the car falls below £1000 for the first time n years after it was purchased.
(b) Find the value of n.(3)
I don't understand how to do part b because I thought the nth term would be n1, so surely it would be 18000*0.8^n1 but it is just 18000*0.8^n in the mark scheme.
However in this question (part b) in order to find the nth term, it uses n1:
A business is expected to have a yearly profit of £275 000 for the year 2016. The profit isexpected to increase by 10% per year, so that the expected yearly profits form a geometricsequence with common ratio 1.1
(a) Show that the difference between the expected profit for the year 2020 and the expectedprofit for the year 2021 is £40 300 to the nearest hundred pounds.(3)
(b) Find the first year for which the expected yearly profit is more than one million pounds.(4)
Thank you for helping.

coconut64
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 16022016 23:50

coconut64
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 16022016 23:51
I just don't get when you would use n and when to use n1...

maggiehodgson
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 17022016 00:19
(Original post by coconut64)
6. A car was purchased for £18 000 on 1st January. On 1st January each following year, the value of the car is 80% of its value on 1st Januaryin the previous year.
(a) Show that the value of the car exactly 3 years after it was purchased is £9216.(1)The value of the car falls below £1000 for the first time n years after it was purchased.
(b) Find the value of n.(3)
I don't understand how to do part b because I thought the nth term would be n1, so surely it would be 18000*0.8^n1 but it is just 18000*0.8^n in the mark scheme.
However in this question (part b) in order to find the nth term, it uses n1:
A business is expected to have a yearly profit of £275 000 for the year 2016. The profit isexpected to increase by 10% per year, so that the expected yearly profits form a geometricsequence with common ratio 1.1
(a) Show that the difference between the expected profit for the year 2020 and the expectedprofit for the year 2021 is £40 300 to the nearest hundred pounds.(3)
(b) Find the first year for which the expected yearly profit is more than one million pounds.(4)
Thank you for helping.
1 year after purchase use n=2 (n1)
2 years after purchase use n=3 (n1)
I think that's it. 
coconut64
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 17022016 00:23
(Original post by maggiehodgson)
In the first question, if you use n, then you are dealing with the nth term but if you want to find it n years after it was bought you use n1.
1 year after purchase use n=2 (n1)
2 years after purchase use n=3 (n1)
I think that's it. 
maggiehodgson
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 17022016 00:31
(Original post by coconut64)
But in the question it says the value of the car falls below £1000 for the first time n years after it was purchased. . So, why isn't n1 used ?Thanks 
coconut64
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 17022016 00:34
(Original post by maggiehodgson)
The question tells you that it fell below £1000 n years after it was bought i.e the (n1)th term will be < £1000 
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 17022016 00:39
(Original post by coconut64)
That's exactly what I did, which is 18000*0.8^n1<1000 but the answer is just 18000*0.8^n<1000
so you and I did it the same way. and we got (n1) log(0.8) = log (1/18)
from which you can find the value of n1 and when you've found it you then have to add one
So perhaps the mark scheme, knowing that you'd have to add one, adds it sooner rather than later. 
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 17022016 00:50
(Original post by maggiehodgson)
OK
so you and I did it the same way. and we got (n1) log(0.8) = log (1/18)
from which you can find the value of n1 and when you've found it you then have to add one
So perhaps the mark scheme, knowing that you'd have to add one, adds it sooner rather than later.
18000* 0.8^n< 1000
and n at the end would be 13 but with my answer n would be 14.. Is it because normally the common ratio is greater than 1 but this is decreasing?? 
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 17022016 01:04
(Original post by coconut64)
Why does that work? The mark scheme just says
18000* 0.8^n< 1000
and n at the end would be 13 but with my answer n would be 14.. Is it because normally the common ratio is greater than 1 but this is decreasing??
n < 13.95
n = 13
Oh, I see what you mean. They actually say 18000* 0.8^n< 1000. Makes no sense to me then.
Which paper is it on and I'll have a look at the rest of it.
As for your second question, what is your question? 
coconut64
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 17022016 01:07
(Original post by maggiehodgson)
Using our method n1 < 12.95....
n < 13.95
n = 13
Oh, I see what you mean. They actually say 18000* 0.8^n< 1000. Makes no sense to me then.
Which paper is it on and I'll have a look at the rest of it.
As for your second question, what is your question?
As for the second question, I was trying to show that they are the same type of question but one uses n1 whereas one doesn't. That is just to clarify my question.. Thanks! 
maggiehodgson
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 17022016 01:13
The mark scheme seems bonkers to me
It starts off with a < sign and 3 lines later changes it to a > sign.
Do you think that might be it? 
coconut64
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 17022016 01:15
(Original post by maggiehodgson)
The mark scheme seems bonkers to me
It starts off with a < sign and 3 lines later changes it to a > sign.
Do you think that might be it?Last edited by coconut64; 17022016 at 01:16. 
16Characters....
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 17022016 01:22
(Original post by coconut64)
No because the inequality sign changes because log0.8 is a negative number. If the number is not greater than 1, you will have to switch the sign. There is a video for it but it doesn't explain why it is just n.. http://www.examsolutions.net/alevel...7&solution=6.2
The first term is £18,000 since this is the original price of the car, after 0 years have elapsed.
The 2nd term is the value of the car after 1 year.
The 3rd term is the value of the car after 2 years...
The (n+1)st term is the value of the car after n years have elapsed. So use your general formulaLast edited by 16Characters....; 17022016 at 01:23. 
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 17022016 01:28
Thanks for that video
He does say that the value AFTER n years will be 18000 * 0.8^n and that's right isn't it.
AND thank you so much about the reminder about a log of a number < 1 is negative. 
coconut64
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 17022016 01:30
(Original post by 16Characters....)
The value of the car each year forms a geometric series with common ratio 0.8.
The first term is £18,000 since this is the original price of the car, after 0 years have elapsed.
The 2nd term is the value of the car after 1 year.
The 3rd term is the value of the car after 2 years...
The (n+1)st term is the value of the car after n years have elapsed. So use your general formula 
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 17022016 01:30
(Original post by maggiehodgson)
Thanks for that video
He does say that the value AFTER n years will be 18000 * 0.8^n and that's right isn't it.
AND thank you so much about the reminder about a log of a number < 1 is negative. 
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 17022016 01:44
(Original post by coconut64)
It is okay! He did but I don't get why though.
After n years it's 18000 * 0.8^n
And the question says after n years it falls below £1000 so should use n. The value is less than £1000 in when n =13 ie after 13 years. 
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 17022016 01:47
(Original post by coconut64)
Where does the 1 come from? I have never seen this equation before because normally I just use n1 to solve this kind of question like the second question I have posted. Thanks.Last edited by 16Characters....; 17022016 at 01:49. 
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 17022016 01:50
(Original post by 16Characters....)
The value of the car each year forms a geometric series with common ratio 0.8.
The first term is £18,000 since this is the original price of the car, after 0 years have elapsed.
The 2nd term is the value of the car after 1 year.
The 3rd term is the value of the car after 2 years...
The (n+1)st term is the value of the car after n years have elapsed. So use your general formula
Brilliant explanation. 
maggiehodgson
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 17022016 01:54
(Original post by coconut64)
6. A car was purchased for £18 000 on 1st January. On 1st January each following year, the value of the car is 80% of its value on 1st Januaryin the previous year.
(a) Show that the value of the car exactly 3 years after it was purchased is £9216.(1)The value of the car falls below £1000 for the first time n years after it was purchased.
(b) Find the value of n.(3)
I don't understand how to do part b because I thought the nth term would be n1, so surely it would be 18000*0.8^n1 but it is just 18000*0.8^n in the mark scheme.
However in this question (part b) in order to find the nth term, it uses n1:
A business is expected to have a yearly profit of £275 000 for the year 2016. The profit isexpected to increase by 10% per year, so that the expected yearly profits form a geometricsequence with common ratio 1.1
(a) Show that the difference between the expected profit for the year 2020 and the expectedprofit for the year 2021 is £40 300 to the nearest hundred pounds.(3)
(b) Find the first year for which the expected yearly profit is more than one million pounds.(4)
Thank you for helping.
So after n years it would be 18000*0.8^n
In the second question the first term is the profit that year not the profit after a year. Does that explain the difference?
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