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    This is a very easy question but dont no how to do it. I have been given a question on sequences and series, it is:
    10
    (3r+2)
    r=5

    now I no how to work it out when r=1 but don't no how to do it when r=5. I no the answer is 147, can somebody please explain how I get to that. Many Thanks
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    just write out the first three term and treat it like a normal arithmetic sequence.
    Be careful about what n is (number of terms)
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    you can rewrite it as

    $

\sum_{r=5}^{10}(3r+2)=

\sum_{r=1}^{10}(3r+2)-\sum_{r=1}^4(3r+2)
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    15
    you need to use the formula:

    0.5n(a+l)
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    the numbers above and below that sign tells you what numbers to put into the equation. i think

    so basically you put the numbers 5-10 into the equations to get 32+29+26+23+20+17
    which equals 147.
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    15
    why is everyone confusing him?

    just use:

    {\frac{1}{2}}n (a + L)

    n=number of numbers in the sequence, which is 6 because you have 5th 6th 7th 8th 9th and 10th values to cater for
    a=first term which you find by putting r=5
    L = last term which you put r =10

    therefore

    n=6
    a=(3(5)+2)
    L=(3(10)+2)

    put that into {\frac{1}{2}}n (a + L)
    the equation you're meant to know! :p:

    excersize 6G 2.d) ?
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    (Original post by Sparrow Hawk)
    the numbers above and below that sign tells you what numbers to put into the equation. i think

    so basically you put the numbers 5-10 into the equations to get 32+29+26+23+20+17
    which equals 147.
    But I cannot do this because I have a exam question that is:
    200
    (5r-2)
    r=5

    So how would I work out this? I only wanted to no that question I gave so i could work out this question myself, but it didnt help. Anybody no how I could work this out?
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    (Original post by neomilan)
    why is everyone confusing him?

    just use:

    {\frac{1}{2}}n (a + L)

    n=number of numbers in the sequence, which is 6 because you have 5th 6th 7th 8th 9th and 10th values to cater for
    a=first term which you find by putting r=5
    L = last term which you put r =10

    therefore

    n=6
    a=(3(5)+2)
    L=(3(10)+2)

    put that into {\frac{1}{2}}n (a + L)
    the equation you're meant to know! :p:

    excersize 6G 2.d) ?
    Perfect got it now
 
 
 
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