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    ∑ (4r+7)
    r=3

    Hello, yes i know it's quite late.

    I'm very confused on how to solve this? i tried solving it the normal way i would if r=1, except by multiplying by 3 first. but the markscheme said it was wrong. (the answer should be 2044) how do i solve this? please explain.
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    Write out the first few terms to get the initial term and the difference term, then use the sum of an arithmetical series, which you can derive easily:

    S = a + (a+d) + .. (a+(n-1)d)
    S = (a+(n-1)d) + (a+(n-2)d) + .. a
    \therefore 2S = n[2a+(n-1)d]
    \therefore S = \frac{n}{2}[2a+(n-1)d]

    What values of a, d and n did you get?
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    (Original post by RogerOxon)
    Write out the first few terms to get the initial term and the difference term, then use the sum of an arithmetical series, which you can derive easily:

    S = a + (a+d) + .. (a+(n-1)d)
    S = (a+(n-1)d) + (a+(n-2)d) + .. a
    \therefore 2S = n[2a+(n-1)d]
    \therefore S = \frac{n}{2}[2a+(n-1)d]

    What values of a, d and n did you get?
    a = 19
    d=4
    n=30

    is that correct??

    i tried using that very same formula before, but to no avail.
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    (Original post by ZiggyStarDust_)
    a = 19
    d=4
    n=30

    is that correct??
    Check your calculation of n.

    However, I don't get the answer that you gave
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    (Original post by RogerOxon)
    Check your calculation of n.

    However, I don't get the answer that you gave
    im worried that my markscheme might be wrong?? 2044 is apparently the answer.

    So N is not 20? i don't understand what n could be then?
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    (Original post by ZiggyStarDust_)
    30
    ∑ (4r+7)
    r=3

    Hello, yes i know it's quite late.

    I'm very confused on how to solve this? i tried solving it the normal way i would if r=1, except by multiplying by 3 first. but the markscheme said it was wrong. (the answer should be 2044) how do i solve this? please explain.
     \sum_{r=3}^{30}=\sum_{r=1}^{30} -\sum_{r=1}^{2}

    because you'd have  (u_1+u_2+u_3+u_4+\cdots +u_3) - (u_1 + u_2)

    can you see you'll be left with u_3 up to u_30 ?
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    r=3..30. If r were 1..30, you'd have 30 terms, so how many do you have for 3..30?
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    solve it like this https://gyazo.com/d401e7fef288e7176a9a7558275e8fdf
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    (Original post by ZiggyStarDust_)
    im worried that my markscheme might be wrong?? 2044 is apparently the answer.
    Sorry - I made a mistake - it is 2044.
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    (Original post by rogeroxon)
    r=3..30. If r were 1..30, you'd have 30 terms, so how many do you have for 3..30?
    o shet its 27 isnt it
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    (Original post by ZiggyStarDust_)
    im worried that my markscheme might be wrong?? 2044 is apparently the answer.

    So N is not 20? i don't understand what n could be then?
    n=28 since the first term is 3 and the last term is 30 (so 28 terms all together).

    Now solve, the answer should be 2044.
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    (Original post by tripplea)
    n=28 since the first term is 3 and the last term is 30 (so 28 terms all together).

    Now solve, the answer should be 2044.
    why would it be 28 as a pose to 27?

    because 30-3 is 27 right? so
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    (Original post by ZiggyStarDust_)
    why would it be 28 as a pose to 27?

    because 30-3 is 27 right? so
    If
    r=1-30 has 30 terms
    then
    r = 2-30 has 29 terms
    r = 3-30 has 28 terms
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    (Original post by tripplea)
    If
    r=1-30 has 30 terms
    then
    r = 2-30 has 29 terms
    r = 3-30 has 28 terms
    thank you thank you thank you

    so much

    you've been very kind and helpful

    its very much appreciated
    you saved my arse

    thank you
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    30
    ∑ (4r+7)
    r=3

    19, 23, 27 - therefore the difference is 4
    a=19, d=4, n=28

    Sn = n/2[2a+(n-1)d]
    28/2[2(19)+(28-1)4)
    14[38+(27)4]
    Sn = 2044

    Hope that's clear enough for you
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    (Original post by ZiggyStarDust_)
    thank you thank you thank you

    so much

    you've been very kind and helpful

    its very much appreciated
    you saved my arse

    thank you
    You're welcome. Do you get it now?
 
 
 

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