Results 1 to 5 of 5

Math Help - Arithmetic Series

  1. #1
    Junior Member
    Joined
    Oct 2005
    Posts
    30

    Arithmetic Series

    Can someone help me on a few questions which I got stuck on and show me how you got the answer?

    1. Evaluate: Sigma sign n on the top, 1 underneath then in the middle 2r-7.

    2. The first term of an aritmetic series is 14. The common difference and the sum of the first n terms are both -6. Find n.

    3. The first three terms of an arithmetic series are 2x+1, 3x+2 and 5x-1. Find x.

    Thanks in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Oct 2005
    Posts
    17

    Smile

    Quote Originally Posted by Confuzzled?
    Can someone help me on a few questions which I got stuck on and show me how you got the answer?

    1. Evaluate: Sigma sign n on the top, 1 underneath then in the middle 2r-7.

    2. The first term of an aritmetic series is 14. The common difference and the sum of the first n terms are both -6. Find n.

    3. The first three terms of an arithmetic series are 2x+1, 3x+2 and 5x-1. Find x.

    Thanks in advance
    Hi there...

    For Qn 1, for typing easeness, i will denote the Sigma sign n on the top, 1 underneath as (S1).

    (S1) (2r-7)
    = (S1) (2r) - (s1) (7)
    = 2 (S1)(r) -7n
    =2 (1/2)(n)(n+1) - 7n
    =n2 + n - 7n
    =n2 - 6n
    =n(n-6)

    For Qn 2, 1st term = a = 14 ; Sn = d = -6

    Sn = n/2(2a + (n-1)d) => eq 1

    Substitute the following values on top into eq 1 and solve for n.

    For Qn 3,

    1st Term = 2x + 1
    2nd Term = 3x + 2
    3rd Term = 5x - 1

    As they are in arithmetic series, they will have a common difference.

    Thus,

    (3x + 2) - (2x + 1) = (5x - 1) - (3x + 2)

    Solve the above equation for x.

    Hope these can be of some help to u.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Arithmetic series:
    an = a1 +(n-1)d
    Sn = (n/2)(a1 +an)

    ------------------
    1. Evaluate: Sigma sign n on the top, 1 underneath then in the middle 2r-7.

    Your "Sigma sign, n on top, 1 underneath" means Summation of the terms from r=1 to r=n. It is the value/sum of the series from r=1 to r=n.
    term = 2r -7
    a1 = 2(1) -7 = -5
    a2 = 2(2) -7 = -3
    a3 = 2(3) -7 = -1
    So, d = 2
    And, an = a1 +(n-1)d = -5 +(n-1)(2) = -5 +2n -2 = 2n -7
    Hence, "Sigma sign, n on top, 1 underneath"[2r -7] = Sn
    = (n/2)[a1 +an]
    = (n/2)[-5 +2n -7]
    = (n/2)[2n -12]
    = n(n-6) ----------------answer.

    -------------
    2. The first term of an aritmetic series is 14. The common difference and the sum of the first n terms are both -6. Find n.

    a1 = 14
    d = -6
    Sn = -6

    Sn = (n/2)[a1 +an] = (n/2)[a1 +(a1 +(n-1)d)] = (n/2)[2a1 +(n-1)d]
    Substitutions,
    -6 = (n/2)[2(14) +(n-1)(-6)]
    -6 = (n/2)[28 -6n +6]
    -6 = (n/2)[34 -6n]
    -6 = n(17 -3n)
    -6 = 17n -3n^2
    3n^2 -17n -6 = 0
    Use the Quadratic Formula to solve for n, or use factoring if possible,
    (3n +1)(n -6) = 0

    3n +1 = 0
    3n = -1
    n = -1/3 -----reject, because n must be a positive integer.

    n -6 = 0
    n = 6 ----------answer.

    -----------------------------
    3. The first three terms of an arithmetic series are 2x+1, 3x+2 and 5x-1. Find x.

    a1 = 2x +1
    a2 = 3x +2
    a3 = 5x -1

    an = a1 +(n-1)d
    a2 = a1 +(2-1)d
    a2 = a1 +d
    3x +2 = (2x +1) +d
    3x +2 -2x -1 = d
    x+1 = d -------------***

    an = a1 +(n-1)d
    a3 = a1 +(3-1)d
    a3 = a1 +2d
    5x -1 = (2x +1) +2(x+1) <----d = x+1
    5x -1 = 2x +1 +2x +2
    5x -1 = 4x +3
    5x -4x = 3 +1
    x = 4 ------------answer.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Oct 2005
    Posts
    30
    Wow thank you so much guys!
    I jus realised i made a mistake, I knew how 2 do the first one, it was another question that I was stuck on. If anyone can help i'll be sooo grateful:
    1. Sigma (n on the top, 6 on the bottom) in the middle: (4-r/2)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2005
    Posts
    17

    Smile

    Quote Originally Posted by Confuzzled?
    Wow thank you so much guys!
    I jus realised i made a mistake, I knew how 2 do the first one, it was another question that I was stuck on. If anyone can help i'll be sooo grateful:
    1. Sigma (n on the top, 6 on the bottom) in the middle: (4-r/2)


    OKie.. no problem... Solving this question requires the same concept as question 1.

    Now u r dealing wif summation of the 1st 6 terms for the sequence (4 - r/2)
    There's two methods u can adopt.

    First method will be by substitution.

    r=1 ; 4 -1/2 = 3 1/2
    r=2 ; 4 -2/2 = 3
    r=3 ; 4 -3/2 = 2 1/2
    r=4 ; 4 -4/2 = 2
    r=5 ; 4 -5/2 = 1 1/2
    r=6 ; 4 -6/2 = 1

    By summation, 1. Sigma (n on the top, 6 on the bottom) in the middle: (4-r/2)
    = 3.5+3+2.5+2+1.5+1
    =13.5

    Method 2:
    S1 (4 -r/2)
    =S1(4) - S1(r/2)
    =4(6) - 1/2 S1(r)
    =24 - 1/2(n/2)(n+1)
    =24 - 1/4(n)(n+1)

    Considering n=6,
    24 - 1/4(6)(6+1)
    =24 - 1.5(7)
    =13.5

    Solving this type of problem will require you to have a basic knowledge on summation of series. So i suggest u read up abit on it before u attempt. Get the basic concepts rite first.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Arithmetic series.
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: December 20th 2010, 05:03 PM
  2. Arithmetic Series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 10th 2010, 02:34 AM
  3. Arithmetic series
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 6th 2010, 12:24 PM
  4. Sum of Arithmetic Series
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 26th 2009, 03:25 AM
  5. Arithmetic Progression or Arithmetic Series Problem
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: October 8th 2009, 01:36 AM

Search Tags


/mathhelpforum @mathhelpforum