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can someone please help answer this question ?? watch

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    i have attempted most of this paper, howver the first question i am clueless. please help Name:  ImageUploadedByStudent Room1390422911.099893.jpg
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    I am unsure if this is GCSE or A level.For A level there is a formula, but for GCSE, I don't think one is taught. Sorry if any of the following is patronising, as you have not specified what part you do not understad.

    The notation means Sum of arithmetic sequence, bottom number is starting term, top is ending term, and the r is what the function, what you do to the term number to give your term..

    So r is just r, so the term = term number. So it is sum of all integers from 1 to 100, which is indeed 5050.

    The First question can be worked out by expressing it as

    (Sum of numbers from 1 to 100) - (Sum of numbers from 1 to 4) = (Sum of numbers from 5 to 100)
    No formulae is really required, just plug in the gaps.
    5050 - (1 + 2 + 3 + 4 =10) = 5040

    The second question is trickier, as the r has changed. The term is now (Term number + 1)
    But you know that there are 96 terms. 100 number from 1 to 100, get rid of 1 to 4, makes 96.
    Each number is + 1, so you could add 96, 1 for every term, to the answer to obtain 5136.

    If you are at A level, or would just like the formula method for ease, just ask.
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    yes. it is a level C1. could you please explain using the formula


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    (Original post by hamzacheema)
    yes. it is a level C1. could you please explain using the formula


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     \sum^n_1r=\frac{1}{2}n(n+1)
    so  \sum^{100}_5r=\sum^{100}_1r-\sum^4_1 r
    and \sum^{100}_5 (r+1)=\sum^{100}_5r+\sum^{100}_5 1=\sum^{100}_5r+96
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    The formula for the sum of an arithmetic sequence is:

    (n/2) x (2a + (n-1)d)

    n is the number of terms
    a is your first term
    d is your 'common difference'

    For example, for the given sequence:

    n is 100
    a is 1, as the first term is 1.
    d is 1. The difference between each term is 1, as the sequence goes 1,2,3...etc.

    (100/2) x (2(1) + (100-1)1)
    50 x (101) = 5050 Indeed.

    Question a can be expressed as

    (Sum of 1 to 100) - (sum of 1 to 4) = (Sum of 5 to 100)
    So using the formula again.
    ((100/2) x (2(1) + (100-1)1)) - ((4/2) x (2(1) + (4-1)1))
    5050 - 10 = 5040.

    Question B is a bit trickier, the function is (1+r) so each term is the term number +1. Can still be expressed as:

    (Sum of 1 to 100) - (sum of 1 to 4) = (Sum of 5 to 100)
    N = 100 N=4
    A= (1+1) = 2 A =2
    D= 1 D = 1

    ((100/2) x (2(2) + (100-1)1)) - ((4/2) x (2(2) + (4-1)1))
    5150 - 14 = 5136
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    You don't need the formula for the sum of an arithmetic sequence. You only need to know what the sum to 100 is. For the first one take teh original sum then subtract 1,2,3,4 and 5 from it.
 
 
 
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