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!

Arithmetic progression Q

  1. Jul 22, 2010 #1
    1. The problem statement, all variables and given/known data
    water fills a tank at a rate of 150 litres during the first hour, 350 litres during the second hour, 550 litres during the 3rd hour and so on. find the number of hours neccesary to fill a rectangular tank 16m x 7m x7m


    2. Relevant equations
    l=a+(n-1)d

    S= n/2 (a+l)

    where:

    l = term
    a = first no in series
    n = nth term
    d = common difference
    S = sum

    3. The attempt at a solution
    volume of tank = 784

    although it doesnt say i assumed 1m(3) is equal to 1000kg = 1000 litres

    therefore water required to fill tank = 784,000 litres

    a = 150

    d = 200

    what i dont understand is in either equation given for arithmetic progression there are 2 unknows?


    the answer is 88.3 hours
     
    Last edited: Jul 22, 2010
  2. jcsd
  3. Jul 22, 2010 #2

    thrill3rnit3

    User Avatar
    Gold Member

    You have 2 equations in 2 unknowns. Do you know how to solve a system of equations?
     
  4. Jul 22, 2010 #3
    not entirely, i attempted to replaced l in the 2nd equation to the answer for l in the first to give:

    S= n/2 (a+(a+(n-1)d))

    but i couldnt seem to get a decent answer from that
     
  5. Jul 22, 2010 #4

    Mark44

    Staff: Mentor

    Substitute for a and d, which you know, and try a few value of n.
     
  6. Jul 22, 2010 #5

    thrill3rnit3

    User Avatar
    Gold Member

    You know a (first term), d (common difference), and S (sum, which is 784,000).

    You're left with 1 variable, n, which is exactly what you're trying to find (number of terms/hours)
     
  7. Jul 22, 2010 #6
    i dont understand sorry, why would i need to substitute values that i already know?
     
  8. Jul 22, 2010 #7
    the problem im having now is although theres only 1 variable (n) theres 2 of them, and i need to get only 1 for the answer.

    this is what i tryed:

    S= n/2 (a+(a+(n-1)d))
    2S = n (150+(150+(n-1)200))
    2S = n (150+150+200n-200)
    2S = n (150+(200n-50)
    2S = n (100+350n)
    2S = 100n+350n2

    not sure if this is the right method or if it is where to go from here?
     
  9. Jul 22, 2010 #8

    thrill3rnit3

    User Avatar
    Gold Member

    S is the sum of the series, so it's 150+350+550+......

    It should total what is necessary to fill the container tank, right? So what do we substitute for S?
     
  10. Jul 22, 2010 #9

    Mark44

    Staff: Mentor

    Mistake in line below.
    This is the right way, but you have a mistake, noted above.

    You know S, so what you have (when you fix the error) is a quadratic equation in n. Hopefully, you know how to solve quadratic equations.
     
  11. Jul 22, 2010 #10
    ahhh brilliant thanks guys.

    changing my mistake i get:

    2 * 784000 = 250n + 200n2

    giving:

    0 = 200n2 + 250n - 1568000

    then applying to a quadratic equation i get 87.921, not quite 88.3 but then im assuming that 1 cubic metre of water = 1000 litres.

    thanks again that confused me a lot more than it should have
     
  12. Jul 22, 2010 #11

    Mark44

    Staff: Mentor

    Looks like what you did is change to a different mistake.

    This line looks fine (except for missing right parenthesis)
    2S = n (150+(200n-50)
    You need only one pair of parentheses, like so:
    2S = n (150+ 200n-50)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Arithmetic progression Q
  1. Arithmetic progression (Replies: 10)

  2. Arithmetic progression (Replies: 2)

  3. Arithmetic Progression (Replies: 1)

  4. Arithmetic progression (Replies: 9)

  5. Arithmetic progression (Replies: 26)

Loading...