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    Well, this maths question is quite hard to do (for me anyway) - i understand whats happening i just cant prove it. I just mess up, anyway:

    A salesman is paid commission of £10 per week for each life insurance policy which he has sold. Each week he sells one new policy so that he is paid £10 commission in the first week, £20 in the second week, £30 in the third etc.
    In the second year the commission increases to £11 per week on new policies sold, although it remains at £10 per week for policies sold in the first year. He continues to see one policy a week.
    [Q] show that he is paid £542 in the second week of his second year.
    ref: Heinman mod maths, P1 , page 185.

    if anyone can help me out that would be great thanks
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    a = 10
    d = 10

    a + 51d = 10 + 510 = 520

    this is what he is paid on the final week of the first year.

    two weeks in, he is paid 520 + 11*2
    because d changes to 11

    which makes 542.
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    ah ye, why did i not recognise it carried on from previous year - thanks.
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    Pretty handy I had the book in front of me already. I did the whole question while I was there.

    wait until you get to question 35 !
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    hehe i did 18,24,32,38 for hwk my maths class is ez and slow - 18 is quite hard too

    mm q35 looks fun :P
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    i sense a challenge... care to post q35?
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    the first three terms in a series are (m-4), (m+2), and (3m +1)

    given that the series is arithmetic, find m

    given that the series is geometric, find the two possible values of m

    ...

    find the sum to infinity of the convergant of the two geometric series'.

    summed up.

    I've finished the first one and I got a fraction. I'll do the rest after dinner. I think it looks more scary than it is.
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    (Original post by mik1a)
    the first three terms in a series are (m-4), (m+2), and (3m +1)

    given that the series is arithmetic, find m

    given that the series is geometric, find the two possible values of m

    ...

    find the sum to infinity of the convergant of the two geometric series'.

    summed up.

    I've finished the first one and I got a fraction. I'll do the rest after dinner. I think it looks more scary than it is.
    a) (m-4)+(3m+1) = 2(m+2)

    => 2m = 7

    => m=7/2

    b) (m-4)(3m+1)=(m+2)²

    => 2m² -15m -8 = 0

    => m = 8 or m = -1/2

    c) [for m=8] r = (10)/(4) = 5/2 >1 => divergent

    [for m=-1/2] r = (3/2)/(-9/2) = -1/3 > -1 => convergent

    sum to infinity = a/(1-r) = (-9/2)/(1+1/3)

    = (-9/2)/(4/3) = -27/8
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    wd, i couldnt even find m
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    Common difference = m+2 - (m-4) = 6, so
    (3m+1) - (m+2) = 6
    ....m = 3.5

    Common ratio = m+2/m-4 = 3m+1/m+2
    Multiply out etc, m = 8, -1/2

    Sum = (m-4)/1-(1/2) = (6-4)/(1/2)
    = 4.
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    You beat me to it Eplaw. We seem to have a difference of opinion for part b, but i got the same equation as u, check ur solutions.
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    (m-4), (m+2), (3m+1)

    Arithmetic:
    ===========
    (m+2)- (m-4)= (3m+1) - (m+2)
    6 = 2m-1
    m=3.5
    =====

    Geometric
    =========
    (m+2)/(m-4)=(3m+1)/(m+2)
    m^2 + 4m + 4 = 3m^2 - 9m - 4
    2m^2 - 15m - 8 = 0
    (2m+1)(m-8)=0
    m=-1/2, m=8
    ===========

    Sum to infinity
    ===============
    take r=(m+2)/(m-4) to ensure the series converges, (|r|<1)
    (m=-0.5)
    r=1.5/-4.5
    =-1/3
    =====

    a0=m-4=-4.5

    the sum to infinity is S=a0(1/(1-r))

    S=a0/(1-r)
    S=-4.5/(1+1/3)
    S=-4.5/(4/3)
    S=-27/8
    =====

    made a mistake earlier. I think that's it now.
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    (Original post by JamesF)
    You beat me to it Eplaw. We seem to have a difference of opinion for part b, but i got the same equation as u, check ur solutions.
    I did my factorisation wrong.
 
 
 
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