An arithmetic progression (AP) is a sequence of numbers where the difference between any two consecutive terms is always the same. This fixed number is called the common difference. For example, in the sequence 2, 5, 8, 11, ..., each number increases by 3. So, the common difference is 3.

On the other hand, a geometric progression (GP) is a sequence where each term is found by multiplying the previous term by the same fixed number, called the common ratio. For example, in the sequence 3, 6, 12, 24, ..., each number is multiplied by 2 to get the next one. So, the common ratio is 2.

The main difference between AP and GP is how the next number is calculated — in AP, we add a fixed number, while in GP, we multiply by a fixed number. Both types of sequences are commonly used in mathematics and real-life situations like calculating interest, analyzing patterns, or solving problems related to time, distance, and growth.

What is Sum of N terms in AP?

Sum of n terms in an arithmetic progression is given by the formula \(S=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\) in which a = first term, n = number of terms and d = common difference. Let us understand this concept in brief by taking an example. Consider a man putting 100 rupees in his daughter’s piggy bank, in such a way that, He deposits 100 rupees on the first birthday, 150 rupees on the second birthday, 200 rupees on the third birthday and he continues in the same way till his daughter turns 21 years old. Hence the amount put in the piggy bank till the daughter’s 21st birthday shall be given as, 100, 150, 200, 250….

To find the total amount of money till the 21st birthday would be the sum of the amount of money deposited for each year.

S = 100 + 150 + 200 + 250 + …..+1100 – equation (1)

This would be a tedious task for each of the numbers up till the 21st birthday,

Let’s reverse equation (1),

S = 1100+ 1050 + 1000+ ……+150 + 100 – equation (2)

Adding equations (1) and (2) we get,

2S = (100 + 1100) + (150+ 1050) + (200 + 1000)…..(100 + 1100).

2S = 1200 + 1200 + 1200 …..(21 times)

2S = 1200 x 21

\(S=\frac{1200\times 21}{2}\)

S = 12600

Hence the total sum of money till the daughter’s 21st birthday is 12600

Such a progression is known as an arithmetic progression and the same technique is used to find the “n”th term in an AP, given by,

a, a+d, a+2d,…..

Hence the sum of first n terms in an AP is given by the formula:

\(S=\frac{n}{2}[2a+(n-1)d]\)

Learn about Harmonic Progression

The formula for the sum of n terms in an arithmetic progression is \(S=\frac{n}{2}[2a+(n-1)d]\)

Where S = sum of the series

n = nth term.

a = first term

d = common difference.

Learn about Properties of Arithmetic Progression

Sum of N terms of an Arithmetic Progression Proof

To find the sum of first n terms of an arithmetic progression,

Let’s consider an arithmetic progression,

a , (a+d) , (a+2d), …..

The nth term of an AP is given by [a + (n-1) d]

“S” denotes the sum of the first n terms of AP.

We have,

S = a + (a+d) + (a+2d) + …..+[a + (n-1) d]. – equation (1)

Rewriting the terms in reverse order,

S = [a + (n-1) d] + [a + (n-2) d] + ……. + (a+d) + a. – equation (2)

On adding equations (1) and (2),

\(2S=\frac{[(a+(n-1)d]+[a+(n-2)d]+\dots\dots.+(a+d)+a)+a+(a+d)+(a+2d)+\dots..+[a+(n-1)d]}{n\ times}\)

\(S=\frac{n}{2}[2a+(n-1)d]\)

Hence the formula for the first n terms of an arithmetic progression is given by,

\(S=\frac{n}{2}[2a+(n-1)d]\)

\(S=\frac{n}{2}[a+a+(n-1)d]\)

\(a_n=a+(n-1)d\)

\(S=\frac{n}{2}(a+a_n)\)

If there are only “n” terms in an arithmetic progression and \(a_n=l\) where “l” is the last term,

\(S=\frac{n}{2}(a+l)\) – this result is useful when the first and the last terms of AP are given and the common difference between them is not given.

Learn about Sum of Infinite GP

Sum of Natural Numbers

Sum of Infinite Arithmetic Progression

In arithmetic progression (AP), each term increases or decreases by a fixed number called the common difference. When we try to find the sum of terms in an AP, we can easily do that if the number of terms is finite. But what happens when the AP has infinite terms?

Let’s understand this:

• In an infinite AP, the terms go on forever without stopping.

• Since each term in an AP grows or reduces steadily by a constant value, the size of the terms keeps getting larger or smaller without bound.

• So, if you try to add up all the terms of an infinite AP, the total keeps increasing endlessly.

For example:
Let’s say the AP is:
5, 10, 15, 20, 25, ...

Here, the numbers keep getting bigger by 5 each time. If you try to add them:
5 + 10 + 15 + 20 + ...,
the sum will grow forever and never settle at a final value.

Unlike a geometric progression (GP), where the terms shrink when the common ratio is less than 1 (like 1, ½, ¼,...), and the sum can reach a finite value, in an AP, there is no such shrinking. The terms don’t become smaller — they keep increasing or decreasing steadily.

Key Points – Sum of First n Terms of AP

1. General Formula for Sum of AP:
   If the first term is a, the common difference is d, and the number of terms is n,
   then the sum of the first n terms is:
   Sn = (n/2) × [2a + (n − 1)d]
2. Alternate Formula Using First and Last Terms:
   If you know the first term (a₁) and the last term (an),
   then:
   Sn = (n/2) × (a₁ + an)
3. Sum of First n Natural Numbers:
   The sum of the first n natural numbers (1, 2, 3, ..., n) is given by:
   Sn = n(n + 1)/2

Sum of n Terms of AP – Tips and Tricks 

To find the sum of the first n terms in an Arithmetic Progression (AP), there are two easy formulas you can use:

1. If you know the first term (a) and the common difference (d), then:
   Sum = n/2 × [2a + (n − 1)d]
2. If you know the first term (a₁) and the last term (aₙ), then:
   Sum = n/2 × (a₁ + aₙ)

Also, if you're adding the first n natural numbers (like 1 + 2 + 3 + ... + n), you can use:
Sum = n(n + 1)/2

Solved Examples Sum of n terms in AP

Example 1: If the sum of the 14 terms of an AP is 1050 and the first term is 10, then what would be the 20th term?

Solution 1: Given data,

Sum of the first 14 terms in an AP = 1050

First term (a) = 10

\(S_{14}=1050\)

n = 14

a = 10

\(S_n=\frac{n}{2}[2a(n-1)d]\)

\(1050=\frac{14}{2}[20+13d]\)

1050 = 140 + 91d.

910 = 91d

d = 10.

Hence,

\(a_{20}=10+(20-1)\times 10=200\)

Hence the 20th of the given arithmetic progression is 200.

Example 2: Find the sum of first “n” natural numbers.

Solution 2: \(S_{n}=1+ 2+3+…..n\)

Here a = 1 (which is the first term)

The difference between two consecutive terms (d) = 1

The last term “l” = n

Hence,

\(S_n=\frac{n}{2}[2a(n-1)d]\)

\(S_n=\frac{n}{2}(n+1)\)

Example 3: Find the sum of the first 24 terms of the list of numbers whose nth term is given by \(a_n=3+2n\)

Solution 3: \(a_n=3+2n\)

\(a_1=3+2=5\)

\(a_2=3+2\ \times 2=7\)

\(a_3=3+2\ \times 3=9\)

Hence the list of numbers will be 5, 7, 9, 11….

Let’s find out the common difference,

7-5 = 2

9-7 = 2

Hence common difference “d” = 2.

Therefore, the sum of the first 24 numbers in the given series is,

\(S_{24}=\frac{24}{2}[2\times 5+(24-1)\times 2]\)

= 672.

Hence the sum of first “24” terms for the given arithmetic sequence is 672.

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