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Math Help - Arithmetic series help

  1. #1
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    Arithmetic series help

    Hi im a bit confused on this question.

    -

    The first 3 terms of an arithmetical progression are 7, 5.9 and 4.8.
    Find
    (i) the common difference
    (ii) the smallest value of n for which the sum to n terms is negative

    -

    I worked out the common difference to be -1.1, which i know is right. Im just a bit stuck on part 2. I know the formula for the sum of terms in an arethmetic series.

    Sum of n = n/2 (2a + (n-1)d)

    n = term ie:5th term
    a = first term, which in this case is 7
    d = common difference which is -1.1

    I tried to solve it by making Sum of n< 0
    I get a wierd answer. Can someone show me where ive gone wrong.
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  2. #2
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    Quote Originally Posted by nugiboy View Post
    Hi im a bit confused on this question.

    -

    The first 3 terms of an arithmetical progression are 7, 5.9 and 4.8.
    Find
    (i) the common difference
    (ii) the smallest value of n for which the sum to n terms is negative

    -

    I worked out the common difference to be -1.1, which i know is right. Im just a bit stuck on part 2. I know the formula for the sum of terms in an arethmetic series.

    Sum of n = n/2 (2a + (n-1)d)

    n = term ie:5th term
    a = first term, which in this case is 7
    d = common difference which is -1.1

    I tried to solve it by making Sum of n< 0
    I get a wierd answer. Can someone show me where ive gone wrong.
    Sum = n/2 (14 + (n - 1)(-1.1)) = n/2 (15.1 - 1.1n).

    You require the smallest, non-negative, integer value of n such that sum < 0:

    The graph of Sum = n/2 (15.1 - 1.1n) has an n-intercept at n = 13.7272. So round 13.7272 UP: n = 14 is the required answer. You should calculate the sum for n = 13 and then for n = 14 as a check.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Sum = n/2 (14 + (n - 1)(-1.1)) = n/2 (15.1 - 1.1n).

    You require the smallest, non-negative, integer value of n such that sum < 0:

    The graph of Sum = n/2 (15.1 - 1.1n) has an n-intercept at n = 13.7272. So round 13.7272 UP: n = 14 is the required answer. You should calculate the sum for n = 13 and then for n = 14 as a check.
    Thanks for the reply. Im a bit confused on how you got the n - intercept as 13.7272 from Sum = n/2 (15.1 - 1.1n). Could you please explain.
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  4. #4
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    Quote Originally Posted by nugiboy View Post
    Thanks for the reply. Im a bit confused on how you got the n - intercept as 13.7272 from Sum = n/2 (15.1 - 1.1n). Could you please explain.
    Thinking instead of the graph of y = x/2 (15.1 - 1.1 x) (where I've relabelled sum as y and n as x) might help. To get the x-intercept, sub y = 0:

    0 = x/2 (15.1 - 1.1 x).

    From the null factor law (it goes by other names but that's the name I usually use) either:

    x/2 = 0 => x = 0, or

    15.1 - 1.1 x = 0 => 15.1 = 1.1 x => x = 15.1/1.1 = 13.7272 (to 4 dp).

    Capisce?
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    Thinking instead of the graph of y = x/2 (15.1 - 1.1 x) (where I've relabelled sum as y and n as x) might help. To get the x-intercept, sub y = 0:

    0 = x/2 (15.1 - 1.1 x).

    From the null factor law (it goes by other names but that's the name I usually use) either:

    x/2 = 0 => x = 0, or

    15.1 - 1.1 x = 0 => 15.1 = 1.1 x => x = 15.1/1.1 = 13.7272 (to 4 dp).

    Capisce?
    Thanks man! I get it now. I think i just got mixed up with the factorising.
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  6. #6
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    Ehew...govind Jup.
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