Sum of N Terms of AP And Arithmetic Progression
Sum of n terms in a sequence can be evaluated only if we know the type of sequence it is. Usually, we consider arithmetic progression, while calculating the sum of n number of terms. In this progression, the common difference between each succeeding term and each preceding term is constant. An example of AP is natural numbers, where the common difference is 1. Therefore, to find the sum of natural numbers, we need to know the formula to find it. Let us discuss here.
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Sum of n terms in AP | n/2[2a + (n – 1)d] |
Sum of natural numbers | n(n+1)/2 |
Sum of square of ‘n’ natural numbers | [n(n+1)(2n+1)]/6 |
Sum of Cube of ‘n’ natural numbers | [n(n+1)/2]^{2} |
Sum of N Terms 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 known 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 n^{th} term as l. The sum of n terms of AP will 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+l)
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]
S_{n} = n(n+1)/2
Hence, this is the formula to calculate sum of ‘n’ natural numbers.
Solved Examples on Sum of n Terms
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
S_{n} = n/2[2a+(n-1)d]
S_{20} =20/2[2×67+(20-1)(-13)]
S_{20}= 10[134 – 247]
S_{20} = -1130
So, the sum of the first 20 terms is -1130.
Solution: The sum of n terms S_{n} = 441
Similarly, S_{n-1}= 356
a = 13
d= n
For an AP, S_{n} = (n/2)[2a+(n-1)d]
Putting n = n-1 in above equation,
l is the last term. It is also denoted by a_{n}_{.} The result obtained is:
S_{n} -S_{n-1} = a_{n}
So, 441-356 = a_{n}
a_{n} = 85 = 13+(n-1)d
Since d=n,
n(n-1) = 72
⇒n^{2 }– n – 72= 0
Solving by factorization method,
n^{2}-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
- 1^{2} + 2^{2} + 3^{2} + 4^{2} + ………. + n^{2}
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 p^{3} – (p – 1)^{3} = 3p^{2} – 3p + 1. In this identity let us put p = 1, 2, 3…. successively, we get
1^{3} – (1 – 1)^{3} = 3(1)^{2} – 3(1) + 1
2^{3} – (2 – 1)^{3} = 3(2)^{2} – 3(2) + 1
3^{3} – (3 – 1)^{3} = 3(3)^{2} – 3(3) + 1
………………………………………..
………………………………………..
………………………………………..
n^{3} – (n – 1)^{3} = 3(n)^{2} – 3(n) + 1
\(\large n^{3}- 0^{3} = 3(1^{2}+ 2^{2}+ 3^{2}+…+n^{2}) – 3 (1+ 2+ 3+ …+ n) + n\) \(\large \Rightarrow n^{3} = 3\sum_{k=1}^{n}(k^{2}) – 3 \sum_{k=1}^{n} (k) + n\) ——————————(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
3\(\large \sum_{p=1}^{n-1}p^{2} = n^{3} + 3 \left (\frac{n(n+1)}{2} \right ) – n\)
This represents the sum of squares of natural numbers using the summation notation. It can be simplified as:
\(\large S_{n} = \sum_{p=1}^{n-1}p^{2} = \frac{1}{3}\left [ n^{3}+ \frac{3n(n+1)}{2} – n \right ] = \frac{1}{6} \left [ 2n^{3} + 3n(n+1) – 2n \right ] = \frac{n(n+1)(2n+1)}{6}\)
Therefore,
Sum of Square of n terms = [n(n+1)(2n+1)]/6 |
Sum of Cubic Series
- 1^{3} + 2^{3} + 3^{3} + 4^{3} + ………. + n^{3}
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} – p^{4} =4p^{3} + 6p^{2} + 4p + 1. In this identity let us put p = 1, 2, 3…. successively, we get
2^{4} – 1^{4} =4(1)^{3} + 6(1)^{2} + 4(1) + 1
3^{4} – 2^{4} =4(2)^{3} + 6(2)^{2} + 4(2) + 1
4^{4} – 3^{4} =4(3)^{3} + 6(3)^{2} + 4(3) + 1
……..………………………………………..
………..……………………………………..
…………..…………………………………..
(n + 1)^{4} – n^{4} =4n^{3} + 6n^{2} + 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:
\(\large S_{n} = \sum_{p=1}^{n} p^{3} = \frac{1}{4} \left [ n^{2} (n+1)^{2} \right ] = \left (\frac{n(n+1)}{2} \right )^{2}\)Therefore, sum of cube of n terms is:
S = [n(n+1)/2]^{2} |
Video Lesson
Summation Formulas
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