Example: Find the sum of :
(i)  the first 1000 positive integers    (ii)  the first n positive integers
Solution :
(i)  Let S = 1 + 2 + 3 + . . . + 1000
Using the formula  $S_n  =n/2{(a+l)}$ for the sum of the first n terms of an AP, we have
$S_{1000}  =1000/2{(1+1000)}$=500X1001=500500
So, the sum of the first 1000 positive integers is 500500.

(ii)  Let Sn  = 1 + 2 + 3 + . . . + n
Here a = 1 and the last term l is n. 
Therefore,  Sn=${n(n+1)}/2$

Example: Find the sum of first 24 terms of the list of numbers whose nth term is given by 
 $a_n  = 3 + 2n$ 

Solution :
As        $a_n  = 3 + 2n$
So, $a_1=3+2(1)=5$ 
So, $a_2=3+2(2)=7$ 
So, $a_3=3+2(3)=9$ 
So, it forms an AP with common difference $d$=7-5=2;
Therefore $S_n$=$n/2$$[2a+(n-1)d]$
          $S_{24}$ = $24/2$$[2(5)+(24-1)2]$
		          =$12[10+46]$
				=672
So, sum of first 24 terms of the list of numbers is 672.

Example: A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh
 year. Assuming that the production increases uniformly by a fixed number every year, find :
(i)  the production in the 1st year                      (ii)  the production in the 10th year
(iii)  the total production in first 7 years

Solution : 
(i)  Since the production increases uniformly by a fixed number every year, the number of TV sets 
manufactured in 1st, 2nd, 3rd, . . ., years will form an AP.
Let us denote the number of TV sets manufactured in the nth year by $a_n$. 
Then, $a_3=600 $ and $a_7=700$
$a_3=a+2d=600$ and $a_7=a+6d=700$
Solving these two equations:
$d=25$ and $a=550$
Therefore, production of TV sets in the first year is 550. 

(ii) Now $a_10=  a + 9d = 550 + 9 × 25 = 775 $
So, production of TV sets in the 10th year is 775.

(iii)Sum of first $n$ terms of AP=$S_n$=$n/2$$[2a+(n-1)d]$
                           $S_7$=$7/2$$[2(550)+(7-1)25]$
						         =4375
Thus, the total production of TV sets in first 7 years is 4375.