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Homework Help: Tough math sequence problem

  1. Aug 17, 2010 #1
    1. The problem statement, all variables and given/known data

    Two finite arithmetic sequences :
    1, 8, 15, 22, ........ , 2003

    2, 13, 24, 34, ....... , 2004

    HOW MANY TERMS THEY HAVE IN COMMON ??

    I don't want the answer !!

    I want to know how to deal with this type of questions ?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 18, 2010 #2

    Mentallic

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    Homework Helper

    Re: Sequences

    Ok so the first sequence

    [tex]T_{n1}=1,8,15,...,2003[/tex]
    [tex]T_{n1}=7n_1-6[/tex] for [tex]1\leq n_1\leq 287[/tex] for [tex]n_1\inZ[/tex]

    [tex]T_{n2}=2,13,24,...,2004[/tex]

    Find the pattern for this one too, then set the patterns equal and simplify to have [tex]n_1=f(n_2)[/tex] and so since n1 is an integer you want fnnd when the function of n2 is an integer for the specified integers n2. You may need to find the pattern of the modulus (that is, if you know what modulus is, in other words, it's the remainder) for the first few integers n2.
     
  4. Aug 18, 2010 #3

    HallsofIvy

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    Science Advisor

    Re: Sequences

    Those are both arithmetic series, the first having common difference 7, the second common difference 11.

    The nth term of an arithmetic sequence with first term [itex]a_0[/itex] and common difference d is [itex]a_n= a_0+ dn[/itex]

    If, for example, you are asked what terms an arithmetic sequence with first term [math]a_0[/math] and common difference d has in common with an arithmetic sequence with first term [math]b_0[/math] and common difference e, you are asked to find number n and m such that [math]a_0+ dn= b_0= em[/itex]. That can be reduced to [itex]dn- em= b_0- a_0[/itex] which is an example of a "Diophantine equation", a general equation of the form ax+ by= c for integers a, b, c and x,y.

    To solve such a thing, first look for x and y such that ax+ by= 1. You can do that by using the "Euclidean division algorithm". For example, to solve 3x+ 8y= 1, note that 3 divides into 8 twice with remainder 2: 8= (2)3+ 2 or 8- (2)3= 2. Now, 2 divides into 3 once with remainder 1: 3= (1)2+ 1 or 3- (1)2= 1. Replacing the "2" in that equation by 8- (2)3 we have 3- (1)(8- (2)3)= (3)3- (1)8= (3)3+ (-1)(8)= 1. That is, x= 3 and y= -1 give a solution.

    Notice that if we add any multiple of 8 to x and subtract that same multiple of 3 from y, we also get a solution: If 3x+ 8y= 1, then 3(x+ 8k)+ 8(y- 3k)= 3x+ 24k+ 8y- 24k= 3x+ 8y= 1 because the two "24k" terms cancel.

    That is, for this problem, x= 3+ 8k and y= -1- 3k are integer solutions to 3x+ 8y= 1 for all integers k.

    To solve, say 3x+ 8y= n for some n other than 1, just multiply x= 3+ 8k and y= -1- 3k by n.
     
  5. Aug 18, 2010 #4
    Re: Sequences

    First of all, you need to find some regularity through the terms already given. One way to proceed is to calculate the subsequent differences. For the first one:
    [tex]
    \begin{array}{ccc}
    1 & & \\
    & 7 & \\
    8 & & 0 \\
    & 7 & \\
    15 & & 0 \\
    & 7 & \\
    22 & & 0
    \end{array}
    [/tex]
    In general, if the differences of order n are all constant, the sequence can be approximated by a polynomial of degree n. In this case [itex]n = 1[/itex]. But, this is called arithmetic progression. The general term for this sequence is:

    [tex]
    a_{n} = a_{1} + d (n - 1) = 1 + 7(n - 1) = 7n - 6
    [/tex]

    [tex]
    1 \le 7n - 6 \le 2003 \Leftrightarrow 7 \le 7n \le 2009 \Leftrightarrow 1 \le n \le 287
    [/tex]

    Can you do a similar analysis for the second sequence?
     
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