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    Can you guys please show your method on how to work this question out, and provide all the formulas you have used.

    A sequence of terms is defined by Un=3 n -2 n>1

    b) The same sequence can also be defined by the recurrence relation
    Un +1=aun +b u1=1
    Where a and b are constants.
    b)Find the value of a and b.

    9. The third term of an arthimetic series is 5 1/2
    The sum of the first four terms of the series is 22 3/4
    a) Show that the first term of the series is 6 1/4 and find the common difference.
    b)Find the number of positive terms in the series.
    c)Hence , find the greatest value of the sum of the first n terms of the series.

    Could you give me the formulas for these questions, and how to work them out.
    Thank you
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    It's up to you to show us what working you've done so far, otherwise we cannot be of help to you.

    We are not here to publish full answers - it doesn't help you.
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    (Original post by marcusmerehay)
    It's up to you to show us what working you've done so far, otherwise we cannot be of help to you.

    We are not here to publish full answers - it doesn't help you.
    I know the answers, as I have the mark scheme.
    Its just that I wanted guidance on the forumlas to use, and when to use them.
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    The first question you posted gives you two equivalent formulae, hence it would make sense to produce some simultaneous equations somehow.

    The second question requires you to use the formulae for an arithmetic sequence, which you should know.

    In part a) you should look to obtain a simultaneous equation, and use the answer you get to find the answer to b) and therefore c).
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    (Original post by Mag_Nificent)
    I know the answers, as I have the mark scheme.
    Its just that I wanted guidance on the forumlas to use, and when to use them.
    To find the sum of an arithmetic series, you use S_n=\frac{n}{2}[2a+(n-1)d], where n is the number of terms, a is the first term and d is the common difference. That is the only formula you'll need to use.
 
 
 
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