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# Sum of First N terms of an Arithmetic Progression

- March 25, 2018
- Posted by: allexamshelps
- Category: Mathematics

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### Sum of First N terms of an Arithmetic Progression (A.P):

##### Before we understand what it is and how it is done, we better understand how the sum of first n Naturals number is calculated.

##### Although we personally believe that the devil is in details, we also believe that God commands to learn in details (precisely) when the purpose is to learn in such a way that the task of memorizing formulae may become redundant (useless), and there are too many formulae in mathematics.

##### For illustration purpose, we will take only first ten natural numbers:

##### 1, 2, 3…10

##### If we add them, we get:

##### 1+2+3+4+5+6+7+8+9+10=55

##### But there is also a formula which can be used to find the sum of very large N in few seconds.

##### \frac { n(n+1) }{ 2 }

##### \frac { 10(10+1) }{ 2 }

##### \frac { 10\times 11 }{ 2 }

##### \frac { 110 }{ 2 }=55

##### So the answer remains the same whether you use the formula or do it manually.

##### This formula was invented by a German Mathematician named Johann Carl Friedrich Gauss when he was asked to find the sum of first 100 natural numbers i.e., from 1 to 100.

##### He did it like this:

##### Firstly, he wrote all the numbers sequentially (one after another):

##### 1, 2, 3, 4, 5, 6, 7, 8, 9, 10…100 .............................(Sequence 1)

##### Then he reversed the order of these numbers. This means he started from 100, 99, 98, 97…1 ...............................(sequence 2)

##### After that, he added both of these sequences.

##### What is a sequence and what is a difference between Sequence and A.P? **learn here.**

##### 1, 2, 3, 4, 5 . . .100

##### + + + + + +

##### 100 99 98 97 96. . . 1

##### =101, =101 =101 =101 = 101 =101

#### So in this way, he got one hundred 101 i.e. 100 x 110

##### 100 X 101 =10100

##### This product, 10100, when divided by 2 gives the sum of first 100 natural numbers which is 5050.

##### \frac { 10100 }{ 2 } =5050

##### In his example, N=100. In our example N=10

##### So, this is how the formula for finding the sum of first natural numbers was invented.

**\frac { n(n+1) }{ 2 }**

##### Similarly, if we are having 6 terms of an A.P with a common difference c.d. then the sum of this A.P can be found using the same method as in the case of the sum of first natural numbers we found above.

##### Firstly write all the six terms of an A.P

###### { a }_{ 1 }, { a }_{ 1 }+c.d, { a }_{ 1 }+2c.d, { a }_{ 1 }+3c.d, { a }_{ 1 }+4c.d, { a }_{ 1 }+5c.d,

##### + + + + + then add the reversed order

###### { a }_{ 1 }+5c.d, { a }_{ 1 }+4c.d, { a }_{ 1 }+3c.d, { a }_{ 1 }+2c.d, { a }_{ 1 }+c.d, { a }_{ 1 }

##### Each time when we add the element of the first row to respective element of the second row we get:

###### ={ 2a }_{ 1 }+5c.d, ={ 2a }_{ 1 }+5c.d, ={ 2a }_{ 1 }+5c.d, ={ 2a }_{ 1 }+5c.d, ={ 2a }_{ 1 }+5c.d, ={ 2a }_{ 1 }+5c.d,

##### So we have found six { 2a }_{ 1 }+5c.d

##### Or

##### 6 x { 2a }_{ 1 }+5c.d,

##### Or

##### 6({ 2a }_{ 1 }+5c.d),

##### To make this the formula for finding the sum of first n natural numbers, we have to divide it by 2.

##### \frac { { 6(2a }_{ 1 }+5c.d) }{ 2 }

##### To interpret it, we can say

**\frac { n(n+1) }{ 2 }=**\frac { n[2a+(n-1)c.d }{ 2 }

##### we can say that

##### n = n and n+1 = 2a+(n-1)c.d (we know how to find it)

##### We had 6 terms in this A.P i.e., n

##### We had a common difference i.e., c.d

##### And we had a first term {a}_{1}

##### Now \frac { n[2a+(n-1)c.d] }{ 2 }

##### N= the number of terms,

##### 2a came because of the addition;

##### (n-1) indicates that if 6 terms are given use 5

##### This 5 or (n-1) is then multiplied by the common difference.

##### The denominator 2 indicates that this sum is an average.

##### \frac { n }{ 2 } [{ a }_{ 1 }{ \quad a }_{ 1 }+(n-1)c.d]

##### Or

##### \frac { n }{ 2 } [{ a }_{ 1 }+{ \quad a }_{ n }]

##### Where

##### { a }_{ n }={ a }_{ 1 }+(n-1)c.d

##### We had learned to make this formula in the last post, **Click here**

### To understand it practically

##### Suppose we are given { a }_{ 1 }=20 that is the first term = 20.

##### The common difference is 2 i.e., c.d=2

##### And the number of terms n=20

##### Find the sum of this A.P

**\frac { n[2a+(n-1)c.d] }{ 2 }**

**{ S }_{ n }=\frac { 20 }{ 2 } [2\times 20+(20-1)2]**

**{ S }_{ n }=\frac { 20 }{ 2 } [2\times 20+19\times 2]**

**{ S }_{ n }=\frac { 20 }{ 2 } [40+38]**

**{ S }_{ n }=\frac { 20 }{ 2 } \times 78**

**{ S }_{ n }=10\times 78=780**

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