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Thread: Divide a line in to multiple section of increasing length according to a ratio

  1. #1
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    Divide a line in to multiple section of increasing length according to a ratio

    hi,

    I want to divide a line segment which is of a fixed length, in to multiple section, where each successive section is greater than preceding section in length by a particular factor 'a' where 'a > 1'.
    There is a also a minimum and maximum limit on the length of the sections.

    for example if I have a line segment of length 10, and I want to divide it in to sections
    where minimum section length is 1 and maximum section length is 3.

    so my question is that is there any way using which I can determine the number of sections 'n' which can probably give a close result to divide a the line with minimum length approximately equal to 1 and maximum length 3, and each section other than the first one is greater than the previous section by a factor a.
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  2. #2
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    Hello gourish

    Welcome to Math Help Forum!
    Quote Originally Posted by gourish View Post
    hi,

    I want to divide a line segment which is of a fixed length, in to multiple section, where each successive section is greater than preceding section in length by a particular factor 'a' where 'a > 1'.
    There is a also a minimum and maximum limit on the length of the sections.

    for example if I have a line segment of length 10, and I want to divide it in to sections
    where minimum section length is 1 and maximum section length is 3.

    so my question is that is there any way using which I can determine the number of sections 'n' which can probably give a close result to divide a the line with minimum length approximately equal to 1 and maximum length 3, and each section other than the first one is greater than the previous section by a factor a.
    There's no way to get exactly the answers you need - unless you just happen to be lucky.

    If the minimum section is of length $\displaystyle \displaystyle p$, the maximum $\displaystyle \displaystyle q$, the common ratio between consecutive sections $\displaystyle \displaystyle a$, the total length $\displaystyle \displaystyle S$ and the number of sections is $\displaystyle \displaystyle n$, then the lengths form a geometric progression. This gives:
    $\displaystyle \displaystyle q = pa^{n-1}$
    and
    $\displaystyle \displaystyle S = \frac{p(a^n-1)}{a-1}$

    Since $\displaystyle \displaystyle n$ has to be an integer, a solution to this pair of simultaneous equations is unlikely to be possible. A spreadsheet will give you some approximate answers. In the case you quote, $\displaystyle \displaystyle p=1,\; S=10$ we get:
    When $\displaystyle \displaystyle n = 5$ and $\displaystyle \displaystyle a=1.352$, $\displaystyle \displaystyle q\approx3.345$
    and
    When $\displaystyle \displaystyle n = 6$ and $\displaystyle \displaystyle a=1.203$, $\displaystyle \displaystyle q\approx2.517$

    Grandad
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