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Math Help - Arithmetic progression problem

  1. #1
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    Arithmetic progression problem

    How many terms at least of the arithmetic progression 1,4,7,10,.............are needed to give a sum greater than 590,starting from the first term of the arithmetic progression?

    please help me
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  2. #2
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    Quote Originally Posted by mastermin346 View Post
    How many terms at least of the arithmetic progression 1,4,7,10,.............are needed to give a sum greater than 590,starting from the first term of the arithmetic progression?
    S_n = \frac{n}{2}(t_1 + t_n) > 590<br />

    where t_n = t_1 + d(n-1)

    solve for the least value of n that makes the sum greater than 590
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  3. #3
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    Quote Originally Posted by skeeter View Post
    S_n = \frac{n}{2}(t_1 + t_n) > 590<br />

    where t_n = t_1 + d(n-1)

    solve for the least value of n that makes the sum greater than 590

    how?? i dont know how to do..
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  4. #4
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    Do as you have been instructed.

    S_n = \frac{n}{2}(t_1 + t_n)

     = \frac{n}{2}[t_1 + t_1 + (n - 1)d]

     = \frac{n}{2}[2t_1 + (n - 1)d]

    You should be able to see that t_1 = 1 and d = 3

    So S_n = \frac{n}{2}[2(1) + 3(n - 1)]

     = \frac{n}{2}(2 + 3n - 3)

     = \frac{n}{2}(3n - 1)

     = \frac{3n^2}{2} - \frac{n}{2}.


    You know that S_n > 590

    So \frac{3n^2}{2} - \frac{n}{2} > 590

    Solve for n.

    Choose the first whole value of n that is greater than what you found...
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    Quote Originally Posted by Prove It View Post
    Do as you have been instructed.

    S_n = \frac{n}{2}(t_1 + t_n)

     = \frac{n}{2}[t_1 + t_1 + (n - 1)d]

     = \frac{n}{2}[2t_1 + (n - 1)d]

    You should be able to see that t_1 = 1 and d = 3

    So S_n = \frac{n}{2}[2(1) + 3(n - 1)]

     = \frac{n}{2}(2 + 3n - 3)

     = \frac{n}{2}(3n - 1)

     = \frac{3n^2}{2} - \frac{n}{2}.


    You know that S_n > 590

    So \frac{3n^2}{2} - \frac{n}{2} > 590

    Solve for n.

    Choose the first whole value of n that is greater than what you found...
    what do you mean by "Choose the first whole value of that is greater than what you found"???
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  6. #6
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    Quote Originally Posted by mastermin346 View Post
    what do you mean by "Choose the first whole value of that is greater than what you found"???
    You have been given almost the entire solution. Your job is to read and process what you've been told and then finish off the problem.

    Solve \frac{3n^2}{2} - \frac{n}{2} = 590 \Rightarrow \frac{3n^2}{2} - \frac{n}{2} - 590 = 0 for n. Choose the value of n that is positive and satisfies \frac{3n^2}{2} - \frac{n}{2} - 590 \geq 0.
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