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Math Help - Arithmetic progression

  1. #1
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    Arithmetic progression

    Greetings,
    I am quite curious about this problem as I have definitely failed to understand my mistake. I have used a similar strategy for an easier problem where q, r and p were known ( p = 6, q = 5, r = 0 if it helps in any way ) and it worked perfectly for that one.
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  2. #2
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    In arithmetic progression sum of the n terms is given by
    Sn = n/2[2a + ( n - 1 )d
    = na + n^2d/2 - nd/2
    Compare this equation with the given equation. You get
    d/2 = p or d = 2p and
    a - d/2 = q or a = p + q.
    In the given equation r may be zero.
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  3. #3
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    Yes, I know of this solution. However, what I fail to understand is why in mine I get that any r does the job whereas in yours, which is also the solution in the textbook, r is required to be 0 only.
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  4. #4
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    Quote Originally Posted by Logic View Post
    Yes, I know of this solution. However, what I fail to understand is why in mine I get that any r does the job whereas in yours, which is also the solution in the textbook, r is required to be 0 only.
    S_3 = S_2 + a_3

    a_3 = S_3 - S_2 = (9p + 3q + r) - (4p + 2q + r) = 5p + q

    Since a_2 = 3p + q we have that d = 2p.

    Your logic correctly worked out that d = 2p - r.

    However, the only way that d can be equal to both 2p - r and 2p is if r = 0.
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