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Math Help - Arithmetic Progression question

  1. #1
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    Arithmetic Progression question

    Hey guys, I'm having problems with this question:

    The sum of the first 100 terms in an arithmetic progression with first term a and common difference d is T. The sum of the first 50 even-numbered terms, i.e. 2nd, 4th, 6th, ..., 100th, is 0.5T + 100. Find the value of a in terms of T. (Answer: a = T/100 - 198)

    Thanks in advance if you could help me with the above question!
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  2. #2
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    Quote Originally Posted by margaritas
    Hey guys, I'm having problems with this question:

    The sum of the first 100 terms in an arithmetic progression with first term a and common difference d is T. The sum of the first 50 even-numbered terms, i.e. 2nd, 4th, 6th, ..., 100th, is 0.5T + 100. Find the value of a in terms of T. (Answer: a = T/100 - 198)

    Thanks in advance if you could help me with the above question!
    You have a sum,
    a+(a+d)+(a+2d)+...+(a+99d)
    This is what I like to do since I hate formulas because it is better to understand than to remember the facts. So I am not going to use the sum formula (in fact I do not even know it!) instead rewrite this like this.
    (\underbrace{a+a+...+a}_{100 \mbox{ times}})+d(1+2+...+99)
    You should know that the sum of the first "n" integers is:
    \frac{n(n+1)}{2}
    Thus,
    100a+d\left( \frac{100(99)}{2} \right)
    Thus,
    T=100a+4950d (1)
    ---
    The second sum is,
    (a+d)+(a+3d)+...+(a+99d)
    I think it is easier to express it in "sigma notation".
    \sum_{k=1}^{50}a+(2k-1)d
    Thus,
    \sum_{k=1}^{50}a+2kd-d
    Use the fact, the sum of the first "n" integers,
    In sigma notation it says,
    \sum_{k=1}^n k=\frac{n(n+1)}{2}
    Thus, in this case,
    50a+2\left( \frac{50(51)}{2} \right) d - 50 d
    Note the 50 in front on the "a" and "d" because you add them 50 times.
    Simplify your expression,
    50a+2500d=\frac{1}{2}T+100 (2)
    Now you have two equation (1) and (2) you should be able to solve for "a" in terms of "T".
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  3. #3
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    Thanks ThePerfectHacker for your solution but I still dont quite get it.

    EDIT: Got the answer, finally! Thanks once again!

    RE-EDIT: No actually, what if I would to use formulas to solve this question? How would the solution be presented?
    Last edited by margaritas; September 1st 2006 at 08:30 AM.
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  4. #4
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    Quote Originally Posted by margaritas
    RE-EDIT: No actually, what if I would to use formulas to solve this question? How would the solution be presented?
    The last two equations that you solve are identical.
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  5. #5
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    Hello, margaritas!

    I had to baby-talk my way through this one.
    See if you like my approach . . .


    The sum of the first 100 terms in an A.P. with first term a and common difference d is T.
    The sum of the first 50 even-numbered terms is \frac{1}{2}T + 100
    Find the value of a in terms of T. .(Answer: a \:= \:\frac{T}{100} - 198)

    We're expected to know the formula for the sum of the first nterms of an A.P.
    . . . S_n\;=\;\frac{n}{2}\left[2a + d(n-1)\right]


    The sum of the first 100 terms is T.
    . . S_{100} \:=\:\frac{100}{2}\left[2a + 99d\right] \:=\:T

    . . which simplifies to: . 100a + 4950d\:=\:T [1]


    The first fifty even terms begin with a_2 = a + d and have a common difference of 2d.

    Hence: . S_{\text{50 even}} \:=\:\frac{50}{2}\left[2(a+d) + 49(2d)\right] \:=\:\frac{1}{2}T + 100

    . . which simplifies to: . 100a + 5000d\:=\:T + 200 [2]


    Subtract [1] from [2]: . 50d = 200\quad\Rightarrow\quad d = 4


    Substitute into [1]: . 100a + 4950(4)\:=\:T\quad\Rightarrow\quad 100a = T - 19800


    Therefore: . \boxed{a\:=\:\frac{T}{100} - 198}

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  6. #6
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    Yay thanks so much Soroban, now I know how to get the second equation
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