Intro, Lecture 2

Sum of First N terms of an Arithmetic Progression

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The Sum of First N terms of an Arithmetic Progression or Arithmetic Sequence: Practical Examples with the simple method of formula derivation.

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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 the first n Naturals number is calculated.

 

Although we personally believe that the devil is in details, we also know 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 the 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 a 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 the 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 using 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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