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Math Help - Help me in solving this problem

  1. #1
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    Help me in solving this problem

    Hi i am new here. I was going through a problem i am unable to understand so i have searched a method online GuruBix - YouTubeCan anyone help me in understanding the problem of Sum of N terms of an Arithmetic Progression or do you have an simpler method of solving Sum of N terms of an Arithmetic Progression. Next month i am going to attend maths exam in my school i request anyone could help me in this ..thanks in advance
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  2. #2
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    Re: Help me in solving this problem

    Quote Originally Posted by kpinky View Post
    Hi i am new here. I was going through a problem i am unable to understand so i have searched a method online GuruBix - YouTubeCan anyone help me in understanding the problem of Sum of N terms of an Arithmetic Progression or do you have an simpler method of solving Sum of N terms of an Arithmetic Progression. Next month i am going to attend maths exam in my school i request anyone could help me in this ..thanks in advance
    Sum of N terms of an Arithmetic Progression
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  3. #3
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    Re: Help me in solving this problem

    The simple way to do a problem of this kind is to recognize that an arithmetic sequence is "linear"- if you were to graph the points (x, y)= (n, a_n), the graph would be a straight line- and linear functions have the property that the "average" of all the numbers is the same as the average of the first and last numbers.

    If you have an arithmetic sequence with "initial value" a_1 and "common difference" d, then the sequence is a_1, a_2= a_1+ d, a_3+ 2d, ... or, generally, a_n= a_1+ (n-1)d.

    In particular, if we were to sum the "i" numbers from a_k to a_{k+i} then the first number in the sequence is [tex]a_k= a_1+ (k-1)d[/tex[] and the last is a_{k+i}= a_1+ (k+i-1)d. The average of those two numbers is \frac{a_k+ a_{k+i}}{2}= \frac{a_1+ (k-1)d+ a_1+ (k+i-1)d}{2}= a_1+ d\frac{2k+ i- 2}{2}= a_1+ d(k- 1+ \frac{i}{2}) so the sum is a_1i+ di(k-1+ \frac{i}{2}).
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  4. #4
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    Re: Help me in solving this problem

    Consider S= 2 + 5 +8 +11 +14+17 In reverse order
    S=17+14+11+ 8 +5 +2 Add both lines together
    2S=19+19+19+19+19 +19=6x19 So S=1/2(6x19)=6/2x19
    Note 6=number of terms and 19 = 1st term + last term
    This will of course work for any arithmetic sequence.
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