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Math Help - Sequences and series help.

  1. #1
    Junior Member
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    Sequences and series help.

    Im stuck on the following questions and need some help. Sorry I dont know how to do sigma and the writing to the right or up or below it so I list it in a clockwise motion from up to right to below.

    Q1 - Calculate Sigma (20, (3+2r), r=1)

    Q2 - The rth term of a series is u. Given that ur = u(r-1) +2 and u1 = 3. Write down u2, u3 and u4 hence find Sigma (n, Ur, r=1)

    Q3 - The 8th term of an arithmetic series is 11 and the 15th term is 39 Find the common difference, the first term in the series and then the nth term.

    Q4 - Find the sum of the intergers which are divisible by 3 and lies between 1 and 400, then find the sum of the intergers from 1 to 400 inclusive which are not divisible by 3.

    Q5 - X(squared), (5x + 1) and (8x - 1) are the first three terms of an arithmetic series. Find the values of x and given that x>0 find the common difference and the first term.

    Q6 - The sequnce u1, u2, u3 is generated from the relation
    u(n+1) = n+1/n Un and U1 = 2 find the values of u2, u3 and u4

    Sorry I asked for so much but I self teach and these are past paper questions that I don't understand, so sorry to cause trouble and if you dont mind can you also write out your workings as well to see how to come to that conclusion so I can handle questions of the smilar sort as future reference.

    Thank You.
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  2. #2
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    Hello, novadragon849!

    Here's some help . . .


    Q3) The 8th term of an arithmetic series is 11 and the 15th term is 39.
    Find the common difference, the first term in the series and then the nth term.
    The nth term is: .a(n) .= .a + (n-1)d
    . . where a is the first term, d is the common difference.

    We have: . a(8) . = -a + 7d . = .11 -[1]
    . . . .and: .a(15) .= .a + 14d .= .39 .[2]

    Subtract [1] from [2]: .7d = 28 . . d = 4

    Substitute into [1]: .a + 28 .= .11 . . a = -17

    Therefore: .a(n) .= .-17 + 4d



    Q4) Find the sum of the intergers which are divisible by 3 and lies between 1 and 400.
    Then find the sum of the intergers from 1 to 400 inclusive which are not divisible by 3.
    The sum of the first n terms of an arithmetic series is given by:
    . . . S(n) .= .n[2a + (n-1)d]

    The sum of the numbers divisible by 3 is: .3 + 6 + 9 + 12 + . . . 399
    . . This is an arithmetic series with: a = 3, d = 3, n = 133

    Its sum is: .S .= .(133)[2(3) + 132(3)] .= .26,733


    The sum of all the numbers from 1 to 400 is: .(400)(401)/2 .= .80,200

    Therefore, the sum of number not divisible by 3 is: .80,200 - 26,733 .= .53,467

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  3. #3
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    oh ok thank you but do you have the answers for the other questions
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