Sum of First n Terms of an AP

Sum of First n Terms of an AP

Let us consider the situation again given in Section 5.1 in which Shakila put   100 into her daughter’s money box when she was one year old, 150 on her second birthday, 200 on her third birthday and will continue in the same way. How much money will be collected in the money box by the time her daughter is 21 years old?

Here, the amount of money (in  ) put in the money box on her first, second, third, fourth . . . birthday were respectively 100, 150, 200, 250, . . . till her 21st birthday. To find the total amount in the money box on her 21st birthday, we will have to write each of the 21 numbers in the list above and then add them up. Don’t you think it would be a tedious and time consuming process? Can we make the process shorter? This would be possible if we can find a method for getting this sum. Let us see.
We  consider  the  problem  given  to  Gauss  (about  whom  you  read  in Chapter 1), to solve when he was just 10 years old. He was asked to find the sum of the positive integers from 1 to 100. He immediately replied that the sum is 5050. Can you guess how did he do? He wrote:
S = 1 + 2 + 3 + . . . + 99 + 100
And then, reversed the numbers to write
S = 100 + 99 + . . . + 3 + 2 + 1

Adding these two, he got
2S = (100 + 1) + (99 + 2) + . . . + (3 + 98) + (2 + 99) + (1 + 100)
= 101 + 101 + . . . + 101 + 101  (100 times)
So,      i.e., the sum = 5050.
We will now use the same technique to find the sum of the first n terms of an AP:
a, a + d, a + 2d, . . .
The nth term of this AP is a + (n – 1) d. Let S denote the sum of the first n terms of the AP. We have
S = a + (a + d) + (a + 2d) + . . . + [a + (n – 1) d]                        (1)
Rewriting the terms in reverse order, we have
= [a + (n – 1) d] + [a + (n – 2) d] + . . . + (a + d) + a                  (2)
On adding (1) and (2), term-wise. We get

or,     2S = n [2a + (n – 1) d]           (Since, there are n terms)

or,   

So, the sum of the first n terms of an AP is given by                   

We can also write this as   

i.e.,                                                       (3)

Now, if there are only n terms in an AP, then a_n = l, the last term. From (3), we see that
                                            (4)