1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

An arithmetic progression problem

  1. Sep 6, 2004 #1
    in an arithmetic sequence there is an even number of term's
    the sum of terms in the odd places is 440 and the sum of terms in the even places is 520, the last term is bigger than the first term by 156
    find how many term's the arithmetic sequence has.
     
  2. jcsd
  3. Sep 6, 2004 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    perhaps you should post this in the homework section?
     
  4. Sep 6, 2004 #3
    I Suppose you know the formula of summation of series:
    [tex]
    S=\frac{n}{2}(2a + (n-1)d)
    [/tex]
    (If not it is easy to derive)
    where [tex]a[/tex]is the first term [tex]n[/tex] is the number of terms and
    [tex]d[/tex] is the common difference between them

    Hint:
    Let the series be:
    [tex]a,a+d,a+2d,.....[/tex]

    even terms [tex]a,a+2d,.....[/tex]
    odd terms [tex]a+d,a+3d,.....[/tex]

    These are Sequences with common diference 2d.
    Use their sum to get 2 eqns
    Using last term - first term = 156 you have 3 eqns 3 unknowns(a,n,d).
    (Need anymore help)
    P.S:Give me Homework Helper medal Quickly PLEEEEEASE!
    EDIT:I gave the wrong formula as I was entranced by the latex graphics
    nobody but halls of ivy saw it i think
    EDIT:Derivation of formula
    [tex]a,a+d,a+2d,.....,a+(n-1)d (i)[/tex]
    invert the above
    [tex]a+(n-1)d,a+(n-2)d,a+2d,.....,a+d,a (ii)[/tex]
    [tex](i)+(ii)=>
    2S=2na + n(n-1)d
    ==>
    s=\frac{n}{2}(2a + (n-1)d)
    [/tex]
    This was proposed by gauss(I think)
     
    Last edited: Sep 6, 2004
  5. Sep 6, 2004 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since this is an arithmetic sequence then, taking a1 as the first term in the sequence, a2= a1+ n, a3= a2+ n= a1+ 2n and, in general, ai= a0+ (i-1)n.

    If there are N numbers in the sequence then the last number is aN= a1+ (N-1)n and so the difference between the first and last terms is
    aN-1= (N-1)n= 156.
    The sum of the even terms is a2+a4+a6+...aN=(a1+n)+(a1+ 3n)+ (a1+ 5n)+ ...+ (a1+ (N-1)n= (N/2)a1+n(1+ 3+ 5+ ...+ (N-1)).

    Now, find a formula for 1+ 3+ 5+ ...+ N-1 so you can get another equation for n and N.
     
  6. Sep 6, 2004 #5
    got it solved thanks for your help :smile:
     
  7. Sep 6, 2004 #6
    I tried out this problem, but i can't seem to go ne where with it.

    For sum of even numbers

    (a+d+a+2d(n-1))*n = 1040<------------ 2an + 2dn^2 -dn = 1040

    For sum of odd numbers

    (a + a +2d(n-1))*n = 880<-------------- 2an + 2dn^2 - 2dn = 880

    solved the two system of equation

    dn = 160

    N = total number = 2n

    Nd = 320

    d(N-1) = 156<----------- d = 164

    so solve for N using Nd = 320 = 320/164 = 1.95.....

    Obviously this is not correct. What did i do wrong here?
     
  8. Sep 7, 2004 #7
    here is my solution i hope its understandable enough

    formula's:
    aN = a1 + dn - d
    sN = (2a1 + d(n - 1))*n/2

    odd:
    s = 440
    d = 2d
    n = n
    aN = aN - d
    a1 = a1

    even:
    s = 520
    d = 2d
    n = n
    aN = aN
    a1 = a1 + d

    general:
    s = 960
    d = d
    n = 2n
    aN = aN
    a1 = a1

    *****************
    aN - a1 = 156
    aN = a1 + 2dn - d
    aN - a1 = 2dn - d
    156 = 2dn - d
    156 = d(2n - 1)
    *****************

    even:
    520 = (2a1 + 2d + 2dn - 2d)*n/2
    520 = (2a1 + 2d + d(2n - 2))*n/2
    520 = (2a1 + 2d + 2dn - 2d)*n/2
    520 = (a1 + d + dn - d)*n
    520 = a1n + dn^2
    a1n = 520 - dn^2
    a1 = (520 - dn^2)/n

    general:
    960 = (2a1 + 2dn - d)*2n/2
    960 = (2a1 + 2dn - d)*n
    960 = (2a1 + d(2n - 1)*n
    960 = 2a1 + 156n
    2a1 = 960 - 156n
    a1 = 480 - 78n

    odd:
    440 = (2a1 + 2dn -2d)*n/2
    440 = (a1 + dn - d)*n
    440 = a1n + dn^2 - dn
    a1n = 440 - dn^2 + dn
    a1 = (440 - dn^2 + dn)/n


    ***************
    440 - dn^2 + dn = 520 - dn^2
    440 + dn = 520
    dn = 80
    d = 80/n
    ***************

    ***************
    156 = d(2n - 1)
    ***************

    156 = 80(2n - 1)/n
    156n = 80(2n - 1)
    156n = 160n - 80
    4n = 80
    n = 20
    2n = 40

    d = 80/n
    d = 80/20
    d = 4
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: An arithmetic progression problem
Loading...