# Sum of N Terms of AP And Arithmetic Progression

A sequence is said to be in a progression if its last term can be represented using a formula. If the co-domain of the function is the set of real numbers, it is called a real sequence, and if it is the set of complex numbers on the other hand, it is called a complex sequence.

## The Sum Formula

The sum of N terms of AP is the sum(addition) of first n terms of the arithmetic sequence. It is equal to n divided by 2 times the sum of twice the first term – ‘a’ and the product of the difference between second and first term-‘d’ also know as common difference, and (n-1), where n is numbers of terms to be added.

Sum of n terms of AP = n/2[2a + (n – 1)d

For example:

• 1, 4, 9, 16, 25, 36, 49 ……….625 represents a sequence of squares of natural numbers till 25.
• 3, 7, 11, 15, 19,………..87 forms another sequence, where each of the terms exceeds the preceding term by 4.

If all the terms of a progression except the first one exceeds the preceding term by a fixed number, then the progression is called arithmetic progression. If a is the first term of a finite AP and d is a common difference, then AP is written as – a, a+d, a+2d, ………, a+(n-1)d.

Note: Before learning how to derive a formula to get the sum of n terms in an AP, try this activity:

• Try to get the sum of the first 100 natural numbers without using any formula.

This question was posed in the same way to one of the great mathematicians, Carl Gauss (1777-1855). He is often referred to be ‘Princeps mathematicorum’ (Latin), meaning ‘the foremost of mathematicians’. At that time, his age was 10 yrs. He came up with the answer to the above problem in a matter of seconds.

The proof for the question can be done using the following way:

• The sum of the number can be represented as

Sum = 1+2+3+4+……………+ 97 + 98 + 99 + 100——————————————– (1)

• Even if the order of the numbers is reversed, their sum remains the same.

Sum = 100 + 99 + 98 + 97 + ………..+ 4 + 3 + 2 + 1—————————————– (2)

Adding equations 1 and 2, we get

•  2 ×  Sum = (100+ 1) + (99+2) + (98+3 )+ (97 +4)+ ………..(4+97)+(3+98)+(2+99)+(1+100)
• 2 × Sum = 101 + 101 + 101 + 101 + ………..(4+97)+(3+98)+(2+99)+(1+100)
• 2 ×Sum = 101 (1 + 1 + 1 + …..100 terms)
• 2 × Sum = 101 (100)
• Sum = {101 × 100}/{2}
• Sum = 5050

Using the above method, sum of numbers like 1000, 10000, etc. can also be calculated.

### Sum of Natural Numbers

 Numbers Sum 1-10 55 1-100 5050 1-1000 500500 1-10000 50005000 1-100000 5000050000 1-1000000 500000500000

The interesting thing is that the above method is applicable to any AP (if the last term of the AP is known). Consider the general form of AP with first term as a, common difference as $d$ and last term i.e. the nth term as l. The sum of n terms of AP will hence be

Sum = a + (a+d) + (a+2d) …… + (l-2d) + (l-d) + l——————– (3)

where l= a+(n-1)d

Writing in reverse order, the sum will still remain same.

Sum =l+(l-d)+(l-2d)..…+(a+2d)+(a+d)+a——————- (4)

Adding equations 3 and 4, we get

2 × Sum =(a+l)+[(a+d)+(l-d)]………+[(l-d)+(a+d)]+(l+a)]

2×  Sum =(a+l)+(a+l)………+(a+l)+(a+l)

2 × Sum = n×(a+l)

⇒ Sum = n/2(a+1)

Substituting the value of l in the previous equation, we get

Sum of n terms of AP = n/2[2a + (n – 1)d

For AP of natural numbers, a = 1 and d = 1, Sum of $n$ terms ($S_n$) of this AP can be found using the formula-

Sn = n/2[2×1+(n-1)1]

Sn = n(n+1)/2

Some examples will enhance the understanding of the topic.

Example 1: If the first term of an AP is 67 and the common difference is -13, find the sum of the first 20 terms.

Solution: Here, a = 67 and d= -13

Sn = n/2[2a+(n-1)d

S20 =20/2[2×67+(20-1)(-13)]

S20= 10[134~-~247]

S20 = -1130

So, the sum of first 20 terms is -1130.

Example 2: The sum of $n$ and ($n~-~1$) terms of an AP is 441 and 356 respectively. If the first term of the AP is 13 and the common difference is equal to the number of terms, find the common difference of the AP.

Solution: The sum of n terms Sn = 441

Similarly, Sn-1= 356

a = 13

d= n

For an AP, Sn = (n/2)[2a+(n-1)d]

Putting n = n-1 in above equation,

l is the last term. It is also denoted by an. The result obtained is

Sn  -Sn-1 = an

So, 441-356 = an

an = 85 = 13+(n-1)d

Since d=n,

n(n-1) = 72

⇒n– n – 72= 0

Solving by factorization method,

n2-9n+8n-72 = 0

(n-9)(n+8)=0

So, n= 9 or -8

Since number of terms can’t be negative,

n= d = 9

### Sum of Square Series

• 12 + 22 + 32 + 42 + ………. + n2

This arithmetic series represents the sum of squares of n natural numbers. Let us try to calculate the sum of this arithmetic series.

To prove this let us consider the identity p3 – (p – 1)3 = 3p2 – 3p + 1. In this identity let us put p = 1, 2, 3…. successively, we get

23 – (1 – 1)3 = 3(1)2 – 3(1) + 1

23 – (2 – 1)3 = 3(2)2 – 3(2) + 1

33 – (3 – 1)3 = 3(3)2 – 3(3) + 1

………………………………………..

………………………………………..

………………………………………..

n3 – (n – 1)3 = 3(n)2 – 3(n) + 1

Adding both the sides of the equation, we get

$\large n^{3}- 0^{3} = 3(1^{2}+ 2^{2}+ 3^{2}+…+n^{2}) – 3 (1+ 2+ 3+ …+ n) + n$

$\large \Rightarrow n^{3} = 3\sum_{k=1}^{n}(k^{2}) – 3 \sum_{k=1}^{n} (k) + n$    ——————————(5)

We have already calculated the sum of n natural numbers as

$\large \sum_{p=1}^{n-1}p = S_{n} = \frac{n(n+1)}{2}$

Substituting this value n equation (5), we get

$\large \sum_{p=1}^{n-1}p^{2} = n^{3} – 3 \left (\frac{n(n+1)}{2} \right ) + n$

This represents the sum of squares of natural numbers using the summation notation. It can be simplified as:

$\large S_{n} = \sum_{p=1}^{n-1}p^{2} = \frac{1}{3}\left [ n^{3}+ \frac{3n(n+1)}{2} – n \right ] = \frac{1}{6} \left [ 2n^{3} + 3n(n+1) – 2n \right ] = \frac{n(n+1)(2n+1)}{6}$

### Sum of Cubic Series

• 13 + 23 + 33 + 43 + ………. + n3

This arithmetic series represents the sum of cubes of n natural numbers. Let us try to calculate the sum of this arithmetic series.

To prove this let us consider the identity (p + 1)4 – p4 =4p3 + 6p2 + 4p + 1. In this identity let us put p = 1, 2, 3…. successively, we get

24 – 14 =4(1)3 + 6(1)2 + 4(1) + 1

34 – 24 =4(2)3 + 6(2)2 + 4(2) + 1

44 – 34 =4(3)3 + 6(3)2 + 4(3) + 1

……..………………………………………..

………..……………………………………..

…………..…………………………………..

(n + 1)4 – n4 =4n3 + 6n2 + 4n + 1

Adding both the sides of the equation, we get

$\large (n+1)^{4} – 1^{4} = 4 (1^{3} + 2^{3} + 3^{3} + …..+ n^{3}) + 6 (1^{2} + 2^{2} + 3^{2} + …+ n^{2}) + 4 (1 + 2+3+4+..+ n) + n$ $\large \Rightarrow n^{3} = 4 \sum_{p = 1}^{n}p^{3} + 6 \sum_{p = 1}^{n}p^{2} + 6 \sum_{p = 1}^{n}p^{p} + n$   ———————(5)

We have already calculated the sum of n natural numbers and sum of squares of n natural numbers as

$\large \sum_{p = 1}^{n}p = S_{n} = \frac{n(n+1)}{2}$, and $\large \sum_{p = 1}^{n}p^{2} = S_{n} = \frac{n(n+1)(2n+1)}{6}$

Substituting these values in equation (5) and simplifying, we get

$\large 4\sum_{p = 1}^{n}p^{3} = n^{4} +4n^{2} + 6n^{2} + 4n – 6\left (\frac{n(n+1)(2n+1)}{6} \right ) -4\left (\frac{n(n+1)}{2} \right ) + n$

This represents the sum of squares of natural numbers using the summation notation. It can be simplified as:

$\large S_{n} = \sum_{p=1}^{n} p^{3} = \frac{1}{4} \left [ n^{2} (n+1)^{2} \right ] = \left (\frac{n(n+1)}{2} \right )^{2}$

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