# Thread: Help me in solving this problem

1. ## 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

2. ## Re: Help me in solving this problem

Originally Posted by kpinky
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

3. ## 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 $a_1i+ di(k-1+ \frac{i}{2})$.

4. ## 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.