Sum of First `n` Terms of an AP
● Let us consider the situation again given in Section 5.1 in which Shakila put Rs. 100 into her daughter’s money box when she was one year old, Rs. 150 on her second birthday, Rs. 200 on her third birthday and will continue in the same way.
● Here, the amount of money (in Rs .) 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, 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, `S = (100 xx 101)/2 = 5050` , 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
`S = [a + (n – 1) d] + [a + (n – 2) d ] + . . . + (a + d) + a` ...........(2)
On adding (1) and (2), term-wise. we get
`2S = ( [ 2a + ( n -1) d ] + [ 2a + ( n-1) d] + ..... + [2a + (n -1) d ] + [2a + ( n-1) d] )/( n text ( times ) )`
or, `2S = n [2a + (n – 1) d ] `(Since, there are n terms)
or, `S = n/2 [2a + (n – 1) d ]`
● So, the sum of the first n terms of an AP is given by
`color{orange}{S = n/2 [ 2a + ( n-1 ) d ]}`
We can also write this as `S = n/2 [ a+a + (n-1 ) d ]`
i.e., `S = n/2 (a + a_n )` ......... (3)
● Now, if there are only n terms in an AP, then `a_n = l`, the last term. From (3), we see that
`color{orange}{S = n/2 (a +l )}` ............(4)
● This form of the result is useful when the first and the last terms of an AP are given and the common difference is not given.
● Now we return to the question that was posed to us in the beginning. The amount of money (in Rs) in the money box of Shakila’s daughter on 1st, 2nd, 3rd, 4th birthday,
. . ., were `100, 150, 200, 250, . . .,` respectively.
● This is an AP. We have to find the total money collected on her 21st birthday, i.e., the sum of the first 21 terms of this AP.
Here, a = 100, d = 50 and n = 21. Using the formula :
● `S= n/2 [2a + (n -1 ) d ]` ,
we have `S = (21)/2 [2 xx 100 + ( 21 -1 ) xx 50 ] = 21/2 [200 + 1000 ]`
` = 21/2 xx 1200 = 12600`
● So, the amount of money collected on her 21st birthday is Rs. 12600.
Hasn’t the use of the formula made it much easier to solve the problem?
● We also use `S_n` in place of S to denote the sum of first n terms of the AP. We write `S_(20)` to denote the sum of the first 20 terms of an AP. The formula for the sum of the first n terms involves four quantities S, a, d and n. If we know any three of them, we can find the fourth
`"Remark :"` The nth term of an AP is the difference of the sum to first n terms and the sum to first (n – 1) terms of it, i.e., `color{orange}{a_n = S_n – S_(n – 1)}`.
● Here, the amount of money (in Rs .) 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, 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, `S = (100 xx 101)/2 = 5050` , 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
`S = [a + (n – 1) d] + [a + (n – 2) d ] + . . . + (a + d) + a` ...........(2)
On adding (1) and (2), term-wise. we get
`2S = ( [ 2a + ( n -1) d ] + [ 2a + ( n-1) d] + ..... + [2a + (n -1) d ] + [2a + ( n-1) d] )/( n text ( times ) )`
or, `2S = n [2a + (n – 1) d ] `(Since, there are n terms)
or, `S = n/2 [2a + (n – 1) d ]`
● So, the sum of the first n terms of an AP is given by
`color{orange}{S = n/2 [ 2a + ( n-1 ) d ]}`
We can also write this as `S = n/2 [ a+a + (n-1 ) d ]`
i.e., `S = n/2 (a + a_n )` ......... (3)
● Now, if there are only n terms in an AP, then `a_n = l`, the last term. From (3), we see that
`color{orange}{S = n/2 (a +l )}` ............(4)
● This form of the result is useful when the first and the last terms of an AP are given and the common difference is not given.
● Now we return to the question that was posed to us in the beginning. The amount of money (in Rs) in the money box of Shakila’s daughter on 1st, 2nd, 3rd, 4th birthday,
. . ., were `100, 150, 200, 250, . . .,` respectively.
● This is an AP. We have to find the total money collected on her 21st birthday, i.e., the sum of the first 21 terms of this AP.
Here, a = 100, d = 50 and n = 21. Using the formula :
● `S= n/2 [2a + (n -1 ) d ]` ,
we have `S = (21)/2 [2 xx 100 + ( 21 -1 ) xx 50 ] = 21/2 [200 + 1000 ]`
` = 21/2 xx 1200 = 12600`
● So, the amount of money collected on her 21st birthday is Rs. 12600.
Hasn’t the use of the formula made it much easier to solve the problem?
● We also use `S_n` in place of S to denote the sum of first n terms of the AP. We write `S_(20)` to denote the sum of the first 20 terms of an AP. The formula for the sum of the first n terms involves four quantities S, a, d and n. If we know any three of them, we can find the fourth
`"Remark :"` The nth term of an AP is the difference of the sum to first n terms and the sum to first (n – 1) terms of it, i.e., `color{orange}{a_n = S_n – S_(n – 1)}`.
