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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Mathematics » Sum of the first n natural numbers

Sum of the first n natural numbers

A concise and precise guide consisting of concepts, formulas, andscientific terms devised to provide the sum of the first n natural numbers study material notes.

Table of Content
  •  

Introduction

The concept of finding the sum of first n numbers is highly used while studying arithmetic progression (or AP). To realize the concept behind finding the sum of first n natural numbers, students need to be familiar with arithmetic progression, the sum of n terms, and how it is derived from these concepts. These study material notes on sum of the first n natural numbers are designed to provide students with accurate and easy-to-understand knowledge on the topic – the sum of first n natural numbers along with examples.

What is an Arithmetic Progression or AP?

  • A sequence, say, a1, a2, a3, ……., an,.….. is called an arithmetic sequence or arithmetic progression if an+1 = an + d, n Є N, where a1 is called the first term of the AP and d is called the common difference of the AP (i.e., the common difference between the consecutive numbers in an AP).
  • Considering the standard form of an AP with the first term as ‘a’ and common difference ‘d’, we have: a, a + d, a + 2d, ….

Then, the general nth term of the AP is: an = a + (n – 1)d.

  • In general terms, an Arithmetic Progression or AP is a sequence where the differences between any two consecutive numbers or terms are the same. For example, sequence 2, 4,6,8, is an AP where the next number is twice its previous. Here, a = 2, d = 2 (i.e., by 4 – 2, 6 – 4), and the nth term can be expressed as 2n.

Properties of an Arithmetic Progression

Some of the simple properties of arithmetic progression one should know are given below.

  • On adding a constant to each term of an AP, the resulting sequence is an AP.
  • On subtracting a constant from each term of an AP, the resulting sequence is an AP.
  • The resulting sequence is an AP multiplying a constant to each term of an AP.
  • The resulting sequence is an AP dividing each term of an AP by a non-zero constant.

Terms and Formulas Used in Arithmetic Progression

Here, the following notations are used for the terms of an AP.

  • a = first term: The first term of an AP is the first term of the sequence forming the arithmetic progression
  • d = common difference: An arithmetic progression is a sequence where all succeeding terms except the first are obtained by adding a fixed number of their preceding ones. This ‘fixed number’ is known as the common difference
  • l = last term: The last term of an AP is the last term of the sequence forming the arithmetic progression.
  • Sn = sum of n terms: The sum of n terms, as the name suggests, is the sum of all terms up to the nth term in an AP.

Some common formulas used in AP are given below.

  • Common difference, d = a2 – a1, where a1 = a, and a2 = a + d.
  • Last or nth term of an AP, l = a + (n – 1)d.
  • Sum of n terms, Sn = n/2 (2a + (n – 1)d).

Derivation of the formula for the sum of n terms

Consider a general sequence a, a + d, a + 2d, …, l. 

From the knowledge gathered about AP, we know:

First-term = a, Common difference = d, and Last term = l

We also know that, l = a + (n – 1)d.

Hence, sum of n terms can be written as:

Sn = a + (a + d) + (a + 2d) + …. + (a + (n – 1)d). – (eq. 1)

It can also be rewritten by starting with the last term and subtracting the common difference from it i.e.,

Sn = l + (l – d) + (l – 2d) + …. + (l – (n – 1)d). – (eq. 2)

On adding eq. 1 and eq. 2, we get:

2Sn = (a + l) + (a + l) + (a + l) + …. + (a + l). 

All d terms are canceled out. Hence,

2Sn = n (a + l)

  • Sn = [n (a + l)]/2

On substituting l = a + (n – 1)d in the above equation, we get:

Sn = n/2 [a + (a + (n – 1)d)]

On simplifying, we get:

Sn = n/2 [2a + (n – 1)d]

Derivation and Formula for Sum of First N Natural Numbers

  • Natural numbers form the sequence: 1, 2, 3, 4, …., n.
  • Let the sum of first n natural numbers be S, where,

S = 1 + 2 + 3 + … + n.

  • On comparing the above sequence of natural numbers with the general form of an AP, we get:
  • a = 1, d = (2-1) = 1, l = n, and Sn = S.
  • We know, Sn = n/2 [2a + (n – 1)d]. 
  • On substituting the values, in the above equation we get:

S = n/2 [(2 x 1) + (n -1)1]

  • S = n/2 [2 + n – 1]
  • S = n/2 (n + 1).
  • The general form of the above equation can be expressed as:

Sn = n/2 (a + l).

 

Conclusion

Through this study material notes on the sum of the first n natural numbers, students have been made familiar with the concepts of arithmetic progression, general terms and common formulas used in arithmetic progression, deriving the formula for the sum of n terms and the derivation and formula of the sum of first n natural numbers. These study material notes on sum of the first n natural numbers are designed to provide students with accurate and easy-to-understand knowledge acting as a precise and concise guide for the topic: Sum of First N Natural Numbers.

faq

Frequently asked questions

Get answers to the most common queries related to the JEE Examination Preparation.

Find the sum of first n natural numbers where the last term is 50.

Ans: We know the formula for the sum of first n natural numbers, S = n/2 (a + l). We also know that in the case of natural ...Read full

State the properties of an arithmetic progression.

Ans: Some properties of arithmetic progression one should know are given below. ...Read full

Derive the formula for the sum of n terms.

Ans: Consider a general sequence a, a + d, a + 2d, …, l.  ...Read full

The sum of first n natural numbers is 325. Find n.

Ans: It is given that sum of first n natural numbers, ...Read full

Ans: We know the formula for the sum of first n natural numbers, S = n/2 (a + l).

We also know that in the case of natural numbers, a =1 and l = n.

  • S = 50/2 (1 + 50)
  • S = 25 * (51)
  • S = 1275.

Hence, the sum of first n natural numbers where the last term is 50 is 1275.

Ans: Some properties of arithmetic progression one should know are given below.

  • On adding a constant to each term of an AP, the resulting sequence is an AP.
  • On subtracting a constant from each term of an AP, the resulting sequence is an AP.
  • On multiplying a constant to each term of an AP, the resulting sequence is an AP.
  • On dividing each term of an AP by a non-zero constant, the resulting sequence is an AP.

Ans: Consider a general sequence a, a + d, a + 2d, …, l. 

From the knowledge gathered about AP, we know:

First-term = a, Common difference = d, and Last term = l

We also know that, l = a + (n – 1)d.

Hence, sum of n terms can be written as:

Sn = a + (a + d) + (a + 2d) + …. + (a + (n – 1)d). – (eq. 1)

It can also be rewritten by starting with the last term and subtracting the common difference from it i.e.,

Sn = l + (l – d) + (l – 2d) + …. + (l – (n – 1)d). – (eq. 2)

On adding eq. 1 and eq. 2, we get:

2Sn = (a + l) + (a + l) + (a + l) + …. + (a + l). 

All d terms are cancelled out. Hence,

2Sn = n (a + l)

  • Sn = [n (a + l)]/2

On substituting l = a + (n – 1)d in the above equation, we get:

Sn = n/2 [a + (a + (n – 1)d)]

On simplifying, we get:

Sn = n/2 [2a + (n – 1)d].

Ans: It is given that sum of first n natural numbers, S = 325.

We know that, for natural numbers, 

S = n/2 (n + 1).

On substituting the value of S as 325 in the above equation, we have:

n/2 (n + 1) = 325

  • n2 + n = 650
  • n2 + n – 650 = 0

We will solve the above quadratic equation using the Sridhar Acharya formula.

  • n = −1 ± √{1 − 4×1×(−650)}/2
  • n = –1 ± √(2601)/2
  • n = (-1 ± 51)/2

We know that numbers of terms cannot be negative.

  • n = (-1 + 51)/2
  • n = 50/2
  • n = 25.

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