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Math Help - Arithmetic Progressions

  1. #1
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    Arithmetic Progressions

    Hi all.

    I'm working on the following problem:

    An arithmetic progression has 14 terms. The sum of the odd-numbered terms is 140 and the sum of the even-numbered terms is 161. Find the common difference of the progression and the 14th term.

    Here's what I've done with it so far...

    A.P.

    n = 14

    T_1 + T_3 + T_5 + . . . + T_13 = 140

    T_2 + T_4 + T_6 + . . . + T_14 = 161

    T_1 + T_2 + . . . + T_14 = 140 + 161 = 301 = S_14

    Now, for an A.P.,

    S_n = n / 2 [ 2 * a + (n - 1) * d ]

    S_14 = 14 / 2 [ 2 * a + (14 - 1) * d ] = 301

    7 [ 2a + 13d ] = 301

    2a + 13d = 43 . . . . . . (i)

    So, I get one equation... Now, I need another equation in terms of a and d so that I can solve them simultaneously to find a and d and then the remaining question will be done...

    Any help will be greatly appreciated.

    Thanks,

    Shahz.
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  2. #2
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    7 odd terms:  S_{odd} = x_1 + x_3 + ...x_{13} = 140
    7 even terms:  S_{even} = x_2 + x_4 + ...x_{14} = 161


    but consider:
     S_{even} - S_{odd} = (x_2 + x_4 + ...x_{14}) - (x_1 + x_3 + ...x_{13} ) = 161 - 140

     S_{even} - S_{odd} = (x_2 + x_4 + ...x_{14}) - (x_1 + x_3 + ...x_{13}  ) = 21


     S_{even} - S_{odd} = (x_2-x_1) + (x_4-x_3) .....+(x_{14}-x_{13}) = 21
    each bracketed expression is teh difference between two consecutive terms of the AP. There are 7 of them in total.

    But you KNOW for an AP that the difference between consectuve terms is just d (the common difference)

     S_{even} - S_{odd} = d + d +.... +d =  21
     S_{even} - S_{odd} = 7d =  21
    That gives you d directly.


    Then , you can use your equation to work out a and then get x_{14} (assuming your equation is right, i didn't check it)
    Last edited by SpringFan25; May 29th 2010 at 04:35 PM.
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  3. #3
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    Quote Originally Posted by SpringFan25 View Post
    There are easier ways to do this:

    7 odd terms:  S_{odd} = x_1 + x_3 + ...x_{13} = 140
    7 even terms:  S_{even} = x_2 + x_4 + ...x_{14} = 161


    but consider:
     S_{even} - S_{odd} = (x_2 + x_4 + ...x_{14}) - (x_1 + x_3 + ...x_{13} ) = 161 - 140

     S_{even} - S_{odd} = (x_2 + x_4 + ...x_{14}) - (x_1 + x_3 + ...x_{13}  ) = 21


     S_{even} - S_{odd} = (x_2-x_1) + (x_4-x_3) .....+(x_{14}-x_{13}) = 21
    each bracketed expression is teh difference between two consecutive terms of the AP. There are 7 of them in total.

    But you KNOW for an AP that the difference between consectuve terms is just d (the common difference)

     S_{even} - S_{odd} = d + d +.... +d =  21
     S_{even} - S_{odd} = 7d =  21
    That gives you d directly.


    Then , you can use your equation to work out x_{14} (assuming your equation is right, i didn't check it)
    Wonderfully elegant solution!! Thank you so much.
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  4. #4
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    Quote Originally Posted by pollardrho06 View Post
    Hi all.

    I'm working on the following problem:

    An arithmetic progression has 14 terms. The sum of the odd-numbered terms is 140 and the sum of the even-numbered terms is 161. Find the common difference of the progression and the 14th term.

    Here's what I've done with it so far...

    A.P.

    n = 14

    T_1 + T_3 + T_5 + . . . + T_13 = 140

    T_2 + T_4 + T_6 + . . . + T_14 = 161

    T_1 + T_2 + . . . + T_14 = 140 + 161 = 301 = S_14

    Now, for an A.P.,

    S_n = n / 2 [ 2 * a + (n - 1) * d ]

    S_14 = 14 / 2 [ 2 * a + (14 - 1) * d ] = 301

    7 [ 2a + 13d ] = 301

    2a + 13d = 43 . . . . . . (i)

    So, I get one equation... Now, I need another equation in terms of a and d so that I can solve them simultaneously to find a and d and then the remaining question will be done...

    Any help will be greatly appreciated.

    Thanks,

    Shahz.
    Alternatively....

    the odd terms are

    a,\ a+2d,\ a+4d,\ ...... 7 terms, 6 multiples of "d"

    the even terms are

    a+d,\ a+3d,\ a+5d,...... 7 terms, 7 multiples of "d"

    then

    7a+(2+4+6+8+10+12)d=140

    7a+42d=140

    and

    7a+(1+3+5+7+9+11+13)d

    7a+49d=161

    the difference is 7d

    7d=21
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  5. #5
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    Quote Originally Posted by Archie Meade View Post
    Alternatively....

    the odd terms are

    a,\ a+2d,\ a+4d,\ ...... 7 terms, 6 multiples of "d"

    the even terms are

    a+d,\ a+3d,\ a+5d,...... 7 terms, 7 multiples of "d"

    then

    7a+(2+4+6+8+10+12)d=140

    7a+42d=140

    and

    7a+(1+3+5+7+9+11+13)d

    7a+49d=161

    the difference is 7d

    7d=21
    Fantastic!! Thanks!!
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  6. #6
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    arithmetric progression

    Quote Originally Posted by pollardrho06 View Post
    Fantastic!! Thanks!!
    Hello shahz,

    I did this differently

    sum of 14 terms = 140+161=301
    using the sum formula it simplifies to 2a +13d=43
    I tried a=1 no go a=2 go 13d=39 d=3

    progression 2,5,8,11,........

    The sum of the odds is 161 not 140 and the sum of the evens is 140 not 161.

    bjh
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  7. #7
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    Quote Originally Posted by bjhopper View Post
    Hello shahz,

    I did this differently

    sum of 14 terms = 140+161=301
    using the sum formula it simplifies to 2a +13d=43
    I tried a=1 no go a=2 go 13d=39 d=3

    progression 2,5,8,11,........

    The sum of the odds is 161 not 140 and the sum of the evens is 140 not 161.

    bjh
    The "odds" are the odd-numbered terms T1, T3, T5 etc
    and the "evens" are the even-numbered terms T2, T4, T6 etc

    not the odd and even values in the sequence itself.
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  8. #8
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    Quote Originally Posted by pollardrho06 View Post
    An arithmetic progression has 14 terms. The sum of the odd-numbered terms is 140 and the sum of the even-numbered terms is 161. Find the common difference of the progression and the 14th term.
    Anything wrong with this "short-cut"? :

    odd: 140 / 7 = 20 (4th odd# term)
    even:161 / 7 = 23 (4th even# term)
    So common diff = 23 - 20 = 3 (or 6 individually)
    And 14th term = 7th even term = 23 + 3(6) = 41
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  9. #9
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    Quote Originally Posted by Wilmer View Post
    Anything wrong with this "short-cut"? :

    odd: 140 / 7 = 20 (4th odd# term)
    even:161 / 7 = 23 (4th even# term)
    So common diff = 23 - 20 = 3 (or 6 individually)
    And 14th term = 7th even term = 23 + 3(6) = 41
    Nothing wrong with it at all!

    Odd numbered terms sum as follows...

    a+(a+2d)+(a+4d)+(a+6d)+(a+8d)+(a+10d)+(a+12d)

    =(a+6d-6d)+(a+6d-4d)+(a+6d-2d)+ (a+6d)+(a+6d+2d)+(a+6d+4d)+(a+6d+6d)

    =(a+6d)+(a+6d)+(a+6d)+(a+6d) +(a+6d)+(a+6d)+(a+6d)+6d-6d+4d-4d+2d-2d

    =7(a+6d)=7(middle\ term)=140

    middle\ term=T_7=a+6d=\frac{140}{7}=20

    Doing the same thing with the even numbered terms discovers "d" and "a".
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