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Arithmetic progression

  1. Dec 11, 2015 #1
    1. The problem statement, all variables and given/known data

    the sum of four integers in A.P is 24 and their product is 945.find them


    2. Relevant equations
    ##(a-d)+a+(a+d)+(a+2*d)=24##
    ##2a+d=12##

    ##(a+d)(a-d)(a)(a+2d)=945##
    ##(a^2-d^2)(a^2+2*a*d)=945##
    3. The attempt at a solution

    there are two equations and two unknowns a(one of the integers) and d(common diff of the A.P)
    but I have trouble manipulating the second(product) equation so that the result of the 1st eq can be used.
    help!!
     
  2. jcsd
  3. Dec 11, 2015 #2

    Mark44

    Staff: Mentor

    Solve for one of the variables above in terms of the other, and then substitute for that variable in the equation below.
     
  4. Dec 11, 2015 #3
    I could have done that but I found a more elegant solution(given at the back with hints which I previously ignored)

    assume the integers
    ##(a-3*d)+(a-d)+(a+d)+(a+3*d)=4*a=24##

    ##(6-3*d)(6-d)(6+d)(6+3*d)=945##

    this is much easier to solve.
    and I think we can come up with such terms when 6 terms are asked for?

    ##(a-5*d)+(a-3*d)+(a-d)+(a+d)+(a+3*d)+(a+5*d)##
     
  5. Dec 11, 2015 #4
    and for even number of terms we can have the common difference d instead of 2d

    for 5 terms

    ##(a-2d)+(a-d)+a+(a+d)+(a+2d)##
     
  6. Dec 11, 2015 #5

    Mark44

    Staff: Mentor

    In the above, you're assuming that the common difference between the integers in the AP is 2d, so you'll need to account for that in your answer.
    Same here.
     
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