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    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 n-1, so surely it would be 18000*0.8^n-1 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 n-1:

    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.
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    I just don't get when you would use n and when to use n-1...
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    (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 n-1, so surely it would be 18000*0.8^n-1 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 n-1:

    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.
    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 n-1.

    1 year after purchase use n=2 (n-1)
    2 years after purchase use n=3 (n-1)

    I think that's it.
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    (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 n-1.

    1 year after purchase use n=2 (n-1)
    2 years after purchase use n=3 (n-1)

    I think that's it.
    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 n-1 used ?Thanks
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    (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 n-1 used ?Thanks
    The question tells you that it fell below £1000 n years after it was bought i.e the (n-1)th term will be < £1000
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    (Original post by maggiehodgson)
    The question tells you that it fell below £1000 n years after it was bought i.e the (n-1)th term will be < £1000
    That's exactly what I did, which is 18000*0.8^n-1<1000 but the answer is just 18000*0.8^n<1000
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    (Original post by coconut64)
    That's exactly what I did, which is 18000*0.8^n-1<1000 but the answer is just 18000*0.8^n<1000
    OK

    so you and I did it the same way. and we got (n-1) log(0.8) = log (1/18)
    from which you can find the value of n-1 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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    (Original post by maggiehodgson)
    OK

    so you and I did it the same way. and we got (n-1) log(0.8) = log (1/18)
    from which you can find the value of n-1 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.
    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??
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    (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??
    Using our method n-1 < 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?
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    (Original post by maggiehodgson)
    Using our method n-1 < 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?
    Thank you! It is the Jan 2010 c2 paper question 6 and here is the mark scheme http://qualifications.pearson.com/co...c_20100218.pdf.

    As for the second question, I was trying to show that they are the same type of question but one uses n-1 whereas one doesn't. That is just to clarify my question.. Thanks!
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    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?
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    (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?
    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/a-level...7&solution=6.2
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    (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/a-level...7&solution=6.2
    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 ar^{(n+1) - 1} = ar^n
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    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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    (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 ar^{(n+1) - 1} = ar^n
    Where does the -1 come from? I have never seen this equation before because normally I just use n-1 to solve this kind of question like the second question I have posted. Thanks.
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    (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.
    It is okay! He did but I don't get why though.
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    (Original post by coconut64)
    It is okay! He did but I don't get why though.
    After 1 year it 18000 * 0.8^1
    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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    (Original post by coconut64)
    Where does the -1 come from? I have never seen this equation before because normally I just use n-1 to solve this kind of question like the second question I have posted. Thanks.
    I think I am misunderstanding your issue here. I will leave you with maggiehodgson so I don't confuse you.
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    (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 ar^{(n+1) - 1} = ar^n
    PRSOM.

    Brilliant explanation.
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    (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 n-1, so surely it would be 18000*0.8^n-1 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 n-1:

    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.
    When you did the first bit of the question you showed that after 3 years the value was 9216 by doing 18000*0.8^3

    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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