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    Hi. I was practicing Maths papers and I came across this question:

    7. A student completes a mathematics course and begins to work through past exampapers. He completes the first paper in 2 hours and the second in 1 hour 54 minutes.
    Assuming that the times he takes to complete successive papers form a geometricsequence,
    (a) find, to the nearest minute, how long he will take to complete the fifth paper, (3)
    (b) show that the total time he takes to complete the first eight papers isapproximately 13 hours 28 minutes, (3)
    (c) find the least number of papers he must work through if he is to complete apaper in less than one hour.

    However this part confused me. I checked the mark scheme and (n) was replaced as (n-1). Why? Can someone explain?

    This is Solomon paper B, C2.

    Thanks in advance!
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    (Original post by greentron6)
    Hi. I was practicing Maths papers and I came across this question:

    7. A student completes a mathematics course and begins to work through past exampapers. He completes the first paper in 2 hours and the second in 1 hour 54 minutes.
    Assuming that the times he takes to complete successive papers form a geometricsequence,
    (a) find, to the nearest minute, how long he will take to complete the fifth paper, (3)
    (b) show that the total time he takes to complete the first eight papers isapproximately 13 hours 28 minutes, (3)
    (c) find the least number of papers he must work through if he is to complete apaper in less than one hour.

    However this part confused me. I checked the mark scheme and (n) was replaced as (n-1). Why? Can someone explain?

    This is Solomon paper B, C2.

    Thanks in advance!
    I'm not entirely sure I know what you mean, perhaps a picture of your working or the mark scheme might help clear things up. . .
    You're looking for u_n=ar^{n-1} such that u_n<1, so there's an n-1 there if that's what you mean.
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    (Original post by joostan)
    I'm not entirely sure I know what you mean, perhaps a picture of your working or the mark scheme might help clear things up. . .
    You're looking for u_n=ar^{n-1} such that u_n<1, so there's an n-1 there if that's what you mean.
    Hi. Thanks for the reply.
    Name:  Untitled.png
Views: 103
Size:  14.6 KB This is what the mark scheme said.
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    (Original post by joostan)
    I'm not entirely sure I know what you mean, perhaps a picture of your working or the mark scheme might help clear things up. . .
    You're looking for u_n=ar^{n-1} such that u_n<1, so there's an n-1 there if that's what you mean.
    Why don't I use the sum of instead considering it says the least amount of papers?
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    (Original post by greentron6)
    Hi. Thanks for the reply.
    Name:  Untitled.png
Views: 103
Size:  14.6 KB This is what the mark scheme said.
    Yes, so as I said.
    Let u_n be the time taken to complete the n^{th} paper.
    I'm assuming you have calculated r=0.95 and you have a=2.
    Plug that into the expression for u_n I gave earlier and you're golden.

    Why don't I use the sum of instead considering it says the least amount of papers?
    The sum??
    That's the time to do multiple papers, and we're interested in the time taken to do 1. . .
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    (Original post by joostan)
    Yes, so as I said.
    Let u_n be the time taken to complete the n^{th} paper.
    I'm assuming you have calculated r=0.95 and you have a=2.
    Plug that into the expression for u_n I gave earlier and you're golden.


    The sum??
    That's the time to do multiple papers, and we're interested in the time taken to do 1. . .
    I understand now. Thank you for your help.
 
 
 
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