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    The first two terms of an arithmetic progression are 16 and 24. Find the least number of terms of the progression which must be taken for their sum to exceed 20000.
    Sol. I got d = 8 from the given information.
    Then I substitued the values into the Sum formula.
    I got the quadratic eq. n^2+3n-5000=0 (Not sure which sign to use)
    Also I how can I solve the eq. to find n? I tried doing factorizing and using quadratic formula but I couldn't get the answer. Please help me out.
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    (Original post by kundanad)
    The first two terms of an arithmetic progression are 16 and 24. Find the least number of terms of the progression which must be taken for their sum to exceed 20000.
    Sol. I got d = 8 from the given information.
    Then I subsitued the values into the Sum formula.
    I got the quadratic eq. n^2+3n-5000=0 (Not sure which sign to use)
    Also I how can I solve the eq. to find n? I tried doing factorizing and quadratic formula but I couldn't get the answer. Please help me out.
    Let me see if I remember how to do this...

     a_1 = 16
     a_2 = 24

     d = a_2 - a_1 = 8

    Whilst solving this I'm rewording the question so it makes more sense for me. I interpret the question as follows, find the smallest value of n for which the sum of  a_1+a_2+a_3+...+a_n is greater than  20000

     \frac{n}{2}(2a_1 + (n-1)d) > 20000
     n(2a_1 + (n-1)d > 40000
     32n + 8n^2 -8n - 40000 > 0
     8n^2 + 24n - 40000 > 0
     n^2 + 3n - 5000 > 0

     n = \frac{-3\pm \sqrt{9-4(1)(-5000)}}{2}

    If you cannot solve that you need to revisit the earlier chapters and do more questions on it. Don't progress onto a further topic until you've mastered the previous
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    (Original post by HelpMe??)
    Let me see if I remember how to do this...

     a_1 = 16
     a_2 = 24

     d = a_2 - a_1 = 8

    Whilst solving this I'm rewording the question so it makes more sense for me. I interpret the question as follows, find the smallest value of n for which the sum of  a_1+a_2+a_3+...+a_n is greater than  20000

     \frac{n}{2}(2a_1 + (n-1)d) > 20000
     n(2a_1 + (n-1)d > 40000
     32n + 8n^2 -8n - 40000 > 0
     8n^2 + 24n - 40000 > 0
     n^2 + 3n - 5000 > 0

     n = \frac{-3\pm \sqrt{9-4(1)(-5000)}}{2}

    If you cannot solve that you need to revisit the earlier chapters and do more questions on it. Don't progress onto a further topic until you've mastered the previous
    Oh thats clear. Thank you so much.
    Btw as I checked the marking scheme it says 'Correct answer from trial and improvement gets full mark.' So how do we get answer by using that method or there is one method only which is shown above?
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    (Original post by kundanad)
    Oh thats clear. Thank you so much.
    Btw as I checked the marking scheme it says 'Correct answer from trial and improvement gets full mark.' So how do we get answer by using that method or there is one method only which is shown above?
    You could use trial and improvement, testing various values of n until you find one that gets the sum to be just below 20000 and one that gets it to be just above. However, the quadratic inequality method is much faster. Simply think of the graph of a quadratic with positive x^2 term. It will be above the x-axis (i.e. the quadratic is > 0) both to the left of the lower root and to the right of the higher root. Now you should be able to find the required conditions on n.
 
 
 
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