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Math Help - Arithmetic progression problem.

  1. #1
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    Arithmetic progression problem.

    If the p^{th} term of an arithmetic progression is \frac{1}{q}and the 9^{th} term = \frac{1}{p},show that (pq)^{th} is 1
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  2. #2
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    Re: Arithmetic progression problem.

    This seems to be false. Take the progression 0, 1/16, 2/16, 3/16, ..., p = 2 and q = 16. Then the second term is 1/16 = 1/q, 9th term is 8/16 = 1/p, but the 32th term is 31/16, not 1.
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  3. #3
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    Re: Arithmetic progression problem.

    but my book state that (Pq) th term is 1 sir?
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  4. #4
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    Re: Arithmetic progression problem.

    Books are not written by the great Authority. They are written by humans.

    Prove it out. An arithmetic progression has a starting value (call it 'a') and a constant difference (call it 'd').

    "the pth term is 1/q"

    a + (p-1)d = 1/q

    "the 9th term is 1/p"

    a + (9-1)d = 1/p

    This leads to p\cdot q = \frac{1}{(a+8d)(a-d)+d}

    p*q is going to have to be an integer. You tell me how this can be and then the implications of the result.
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  5. #5
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    Re: Arithmetic progression problem.

    Hello, mastermin346!

    I believe there is a typo . . . It is not the 9th term.


    . . . . . i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . \downarrow
    \text{If the }p^{th}\text{ term of an arithmetic progression is }\tfrac{1}{q}\,\text{ and the }q^{th}\text{ term is }\tfrac{1}{p}
    \text{show that the }(pq)^{th}\text{ term is }1.

    \text{Note that: }\,p \ne q

    \text{The }n^{th}\text{ term is: }\:a_n \:=\:a + (n-1)d
    . . \text{where }a\text{ = first term, }d\text{ = common difference.}

    \text{We are told: }\:\begin{Bmatrix} a_p \:=\:\frac{1}{q} & \Rightarrow & a + (p-1)d &=& \frac{1}{q} & [1] \\ \\[-3mm] a_q \:=\:\frac{1}{p} & \Rightarrow & a + (q-1)d &=& \frac{1}{p} & [2] \end{Bmatrix}

    . . \begin{array}{ccccc}\text{From [1]:} & aq + dpq - dq \:=\:1 & [3] \\ \\[-3mm] \text{From [2]:} & ap + dpq - dp \:=\:1 & [4] \end{array}


    \text{Equate [3] and [4]: }\:aq + dpq - dq \:=\:ap + dpq - dp

    . . ap - aq - dp + dq \:=\:0 \quad\Rightarrow\quad a(p-q) - d(p-q) \:=\:0

    . . (p-q)(a-d)\:=\:0 \quad\Rightarrow\quad \rlap{/////}p = q,\;a = d\;\;[5]


    \text{Substitute [5] into [3]: }\:dq + dpq - dq \:=\:1 \quad\Rightarrow\quad dpq \:=\:1 \quad\Rightarrow\quad pq \:=\:\tfrac{1}{d}

    \text{Then: }\:a_{pq} \;=\;a_{\frac{1}{d}} \;=\;d + \left(\tfrac{1}{d}-1\right)d \;=\; d + 1 - d

    . . \text{Therefore: }\:a_{pq} \;=\;1

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  6. #6
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    Re: Arithmetic progression problem.

    Note that: p\ne q
    This has to be an assumption in the problem statement because nothing prevents pth term to be 1/p; this does not imply that p^2th term is 1.
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