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    Hai,

    I've totally forgtten how to do these if anyone would care to explain

    9c) The rth term of an arithmetic series is (2r-5)

    Show that E (weird symbole with "n" on top of the "E" and r=1 under it)

    E (2r - 5) = n(n-4)

    It's the C1 book page 144 question 9c

    Also, question 11b)

    A sequence of a1, a2, a3 is defined by a1 = 3

    3a + 5 was the formula and i figured out

    a1 = 3 a2 = 4 a3 = 7

    Caculate the value of E (weird symbole with "5" on top and r=1 under it)


    - Thanks
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    (Original post by Sadbunny11)
    Hai,

    I've totally forgtten how to do these if anyone would care to explain

    9c) The rth term of an arithmetic series is (2r-5)

    Show that E (weird symbole with "n" on top of the "E" and r=1 under it)

    E (2r - 5) = n(n-4)

    It's the C1 book page 144 question 9c

    Also, question 11b)

    A sequence of a1, a2, a3 is defined by a1 = 3

    3a + 5 was the formula and i figured out

    a1 = 3 a2 = 4 a3 = 7

    Caculate the value of E (weird symbole with "5" on top and r=1 under it)


    - Thanks
    Your E (weird symbol with "n" on top and r = 1 under) is a sigma. It denotes the sum of a set of terms in the series from r = 1 to n.

    1) The sum of n terms in an arithmetic series is given by  \frac{n}{2}(2r_1 + (n-1)d) , r_1 being the first term, and d the common difference.

    2) Again you're being asked to calculate the sum of the first n terms of a series. I'd point out here that your a1, a2 and a3 don't have a common difference, so I'd look at recalculating those.
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    The E is called sigma
    • Thread Starter
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    Hai - thanks for the reply.

    I just noticed that for 11b) the one that hasn't got a common difference it hasn't actually stated its a arithmetic series (i just assumed it D: - my mistake).

    Is the only way to add each step up then? Since i dont believe i was given a formula.

    Also, thanks for clearing up 9c) aswell, but how would i go from...

    n/2 [2r + (n-1) d]

    to

    (2r-5) = n(n-4)

    A tad confused on that

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    (Original post by Sadbunny11)
    Also, thanks for clearing up 9c) aswell, but how would i go from...

    n/2 [2r + (n-1) d]

    to

    (2r-5) = n(n-4)

    A tad confused on that

    - Thanks
    You aren't trying to equate (2r-5) to n(n-4). You're saying that the sum of the first N terms in the series (2r-5) is equal to n(n-4)

    You have the equation for the sum of terms in any arithmetic sequence (n/2 [2r + (n-1) d]).

    Basically,

    Sigma (2r-5) = n/2 [2r + (n-1) d]
    Get your common difference d, and your first term r, and you're sorted with a bit of simplifying.

    EDIT:

    For the other question, what is actually given to you in the question? The formula?
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    Thankyou!
 
 
 
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