Sum of an Arithmetic Progression: Formula & Examples

The sum of an arithmetic progression (AP) is a fundamental concept in mathematics that helps in finding the total of a sequence with a constant difference between consecutive terms. It is built on the basics of arithmetic sequences and helps calculate sums without adding each term individually. Understanding this concept is essential for solving numerical problems involving patterns, series, and real-life situations. In this guide, you will learn about the sum of an arithmetic progression, its formulas, and real-life applications through solved examples.

Table of Contents

• What is an Arithmetic Progression
• Sum of an Arithmetic Progression
• Derivation of Sum of an Arithmetic Progression
• Special Cases of Sum of an Arithmetic Progression
• Solved Examples on Sum of an Arithmetic Progression
  
  

What is an Arithmetic Progression?

An arithmetic sequence (AP) is a sequence of numbers in which the difference between consecutive terms is always constant. This constant value is called the common difference (d).

General Form: a,a+d,a+2d,a+3d,…

Where: a = first term and d = common difference

Read more: Important Questions on Arithmetic Progression - Class 10

Sum of an Arithmetic Progression

The sum of an arithmetic sequence is the sum of the terms of the sequence up to a certain number of terms. Instead of adding each term individually, we use formulas to calculate the sum quickly and efficiently. This is especially useful when dealing with large sequences.

Formula 1: (when the last term is known):

Sn=n2(a+l)

Where:

• Sn = sum of n terms
• a = first term
• l = last term
• n = number of terms
  
  

Formula 2: (when last term is unknown):

Sn=n2[2a+(n−1)d]

Where:

• Sn = sum of n terms
• a = first term
• n = number of terms
• d = common difference
  
  

Derivation of Sum of an Arithmetic Progression

To find the sum of n terms of an AP: a,a+d,a+2d,a+3d,…

The nth term of the AP is denoted by an=a+(n−1)d. Let S denote the sum of the first n terms of the AP.

S=a+(a+d)+(a+2d)+(a+3d)+⋯+[a+(n−1)d]⋯(1)

Writing the numbers in reverse order, we have:

S=[a+(n−1)d]+[a+(n−2)d]+⋯+(a+d)+a⋯(2)

Adding (1) and (2) termwise, we get:

2S=n[2a+(n−1)d]

S=n2[2a+(n−1)d]

This can be rewritten as:

S=n2[a+an]

If an=l, then:

S=n2(a+l)

Special Cases of Sum of an Arithmetic Progression

Here are a few special cases that make solving AP problems faster and more efficient:

Special Case	Formula
Sum of the first n natural numbers	n(n+1)2
Sum of squares of the first n natural numbers	n(n+1)(2n+1)6
Sum of cubes of the first n natural numbers	[n(n+1)2]2

Solved Examples on Sum of an Arithmetic Progression

Example 1: Find the sum of: 5 + 10 + 15 + ... + 50

Solution: Given a=5, d=5 and l=50

an=a+(n−1)d

50=5+(n−1)×5

n−1=455=9⟹n=10

S=n2(a+l)=102(5+50)=275

Example 2: How many terms of the AP: 3, 7, 11, ... must be taken so that their sum is 406?

Solution: Given a=3, d=4 and Sn=406

Sn=n2[2a+(n−1)d]

406=n2[2(3)+(n−1)4]

406=n2[4n+2]=n(2n+1)

2n2+n−406=0

(n−14)(2n+29)=0

n=14orn=−292

Hence, the number of terms is n = 14.

Example 3: If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.

Solution: Given a=10, n=14 and S14=1050

Sn=n2[2a+(n−1)d]

1050=142[2(10)+(14−1)d]

1050=7[20+13d]⟹d=10

a20=10+(20−1)×10=200

Example 4: Find the sum of the first 25 natural numbers.

Solution: The given AP is 1,2,3,…,25

Given a=1, l=25 and d=1

{Sum of first }n{ natural numbers}=n(n+1)2

{Sum of first 25 terms}=25×262=25×13=325