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Series, arithmatic progression.

  1. Feb 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi All,
    I found this problem,

    The sum of p, q, r terms of an Arithmetic Progression, are P, Q, R respectively: show that


    [tex] \frac{P (q - r )}{p} + \frac{Q (r - p )}{q} + \frac{R (p - q)}{r} = 0 [/tex]

    2. Relevant equations

    3. The attempt at a solution


    My thoughts on how to start the problem is;

    if
    [tex] S_{n} = \frac{a}{2} (n + (n-1)d ) [/tex]

    then the sum of say 'p' terms, would be

    [tex] P = S_{p} = \frac{a}{2} (p + (p-1)d ) [/tex]

    Therefore;

    [tex] Q = S_{q} = \frac{a}{2} (q + (q-1)d ) [/tex]


    [tex] R = S_{r} = \frac{a}{2} (r + (r-1)d ) [/tex]

    If I used the following series, to simplify a little, [tex] S_{n} = 1 + 2 + 3 ... + n, [/tex] then [tex] S_{n} = \frac{1}{2}n(n+1) [/tex]

    But how to form the above equation, which combines all series, which includes all terms (i.e p q, r, P, Q & R)?
     
  2. jcsd
  3. Feb 15, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    I guess you can just plug in those expressions for the sums and simplify.
     
  4. Feb 17, 2015 #3
    Solved it!

    [tex] \frac{P}{p} = \frac{1}{2}(a + d(p-1)) [/tex] [tex] (eq 1) [/tex]

    Use sum of Q, to let a be the subject.

    [tex] a = 2\frac{Q}{q} - d(q-1) [/tex] [tex] (eq 2) [/tex]

    Sub, eq 1 into eq 2

    [tex] d = 2 \frac{Q}{(p-q)} ( \frac{P}{p} - \frac{Q}{q} ) [/tex]

    Do the same for Sum Q & R, solve for d.

    [tex] d = 2 \frac{Q}{(p-r)} ( \frac{Q}{q} - \frac{R}{r} ) [/tex]

    [tex] 2 \frac{Q}{(p-q)} ( \frac{P}{p} - \frac{Q}{q} ) = 2 \frac{Q}{(p-r)} ( \frac{Q}{q} - \frac{R}{r} ) [/tex]

    Simplify,

    [tex] \frac{P (q - r )}{p} + \frac{Q (r - p )}{q} + \frac{R (p - q)}{r} = 0 [/tex]
     
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