A sequence is a group of numbers arranged in a particular order or following a set of rules and a summation of terms in a sequence is called a series. A sequence of numbers where the terms follow a specific pattern is also called a progression. There are many types of numeric progressions such as arithmetic progression, geometric progression, harmonic progression, etc.

Letâ€™s understand what is arithmetic progression and what the different formulas used to solve problems are based on it.

## What is Arithmetic Progression?

An arithmetic progression abbreviated as AP is a sequence of numbers where the difference between every two consecutive terms is the same. In an AP, each term after the first term is obtained by adding a fixed number to its previous term. This fixed number is known as the common difference and is denoted by $d$. The first term of an arithmetic progression is denoted by $a$. Sometimes the first term is also denoted by $a_1$.

For example, $1, 4, 7, 10, 13, 16$, … is an arithmetic progression as the differences between every two consecutive terms are the same (i.e., $3$). $4 – 1 = 7 – 4 = 10 – 7 = 13 – 10 = 16 – 13 = â€¦ = 3$. We can also notice that every term (except the first term) of this AP is obtained by adding $3$ to its previous term.

Thus, an arithmetic progression, in general, can be written as $a$, $a + d$, $a + 2d$, $a + 3d$, â€¦, where $a$ is a first term and $d$ is a common difference.

### Examples of Arithmetic Progression

**Example 1:** Â Meenu has been offered a job with a starting monthly salary of â‚¹$12,000$, with an annual increment of â‚¹$800$ in her salary. Her salary (in â‚¹) for the $1^{st}$, $2^{nd}$, $3^{rd}$, . . . years will be, respectively $12,000$, $12,800$, $13,600$, â€¦.

The values $12,000$, $12,800$, $13,600$, â€¦. are in AP, where the first term $a = 12000$, and common difference $d = 12800 – 12000 = 13600 – 12800 = 800$.

**Example 2:** A city taxi service has a fare structure of â‚¹$15$ for the first km and â‚¹$8$ for each additional km.

Taxi fare for the $1$ km = â‚¹$15$

Taxi fare for the $2$ kms = $15 + 8 =$â‚¹$23$

Taxi fare for the $3$ kms = $23 + 8 =$â‚¹$31$

Taxi fare for the $4$ kms = $31 + 8 =$â‚¹$39$

The fare amounts $15$, $23$, $31$, $39$, â€¦ are in AP, where the first term $a = 15$, and common difference $23 – 15 = 31 – 23 = 39 – 31 = 8$.

**Example 3:** The amount of air present in a cylinder when a vacuum pump removes $\frac{1}{4}$ of the air remaining in the cylinder at a time. Check whether the amount of air present in the cylinder after each time the vacuum pump is used is in AP or not.

Letâ€™s consider that the initial amount of air in a cylinder = $100$ cubic units

Amount of air removed by vacuum pump after first time = $\frac{1}{4} \times 100 = 25$. Amount of air left in the cylinder = $100 – 25 = 75$ cubic units.

Amount of air removed by vacuum pump after second time = $\frac{1}{4} \times 75 = 18.75$. Amount of air left in the cylinder = $75 – 18.75 = 56.25$ cubic units.

Amount of air removed by vacuum pump after third time = $\frac{1}{4} \times 56.25 = 14.0625$. Amount of air left in the cylinder = $56.25 – 14.0625 = 42.1875$ cubic units.

The amount of air in the cylinder(in cubic units) after each use of the vacuum pump is $100$, $75$, $56.25$, $42.1875$, â€¦ which is not an AP.

Difference between $1^{st}$ and $2^{nd}$ terms is $75 – 100 = -25$

Difference between $2^{nd}$ and $3^{rd}$ terms is $56.25 – 75 = -18.75$

Difference between $3^{rd}$ and $4^{th}$ terms is $42.1875 – 56.25 = -14.0625$

The difference between consecutive terms of the sequence is not the same.

**Example 4: **Form an AP whose first term is $12$ and the common difference is $7$.

Here $a = 12$ and $d = 7$.

$1^{st}$ term = $a = 12$

$2^{nd}$ term = $a + d = 12 + 7 = 19$

$3^{rd}$ term = $a + 2d = 12 + 2 \times 7 = 12 + 14 = 26$

$4^{th}$ term = $a + 3d = 12 + 3 \times 7 = 12 + 21 = 33$

$5^{rd}$ term = $a + 4d = 12 + 4 \times 7 = 12 + 28 = 40$

Therefore, the AP is $12, 19, 26, 33, 40, â€¦$

**Example 5: **Form an AP whose first term is $57$ and the common difference is $-11$.

Here $a = 57$ and $d = -11$.

$1^{st}$ term = $a = 57$

$2^{nd}$ term = $a + d = 57 + (-11) = 46$

$3^{rd}$ term = $a + 2d = 57 + 2 \times (-11) = 57 – 22 = 35$

$4^{th}$ term = $a + 3d = 57 + 3 \times (-11) = 57 – 33 = 24$

$5^{rth}$ term = $a + 4d = 57 + 4 \times (-11) = 57 – 44 = 13$

$6^{th}$ term = $a + 5d = 57 + 5 \times (-11) = 57 – 55 = 2$

$7^{th}$ term = $a + 6d = 57 + 6 \times (-11) = 57 – 66 = -9$

Therefore, the AP is $57, 46, 35, 24, 13, 2, -9, â€¦$

**Note: **

- An AP is an increasing AP, when the common difference $d$ is positive
- An AP is a decreasing AP, when the common difference $d$ is negative
- An AP is a constant AP, when the common difference $d$ is zero

## nth Term of Arithmetic Progression

The general term (or) $n^{th}$ term of an AP whose first term is $a$ and the common difference is $d$ is given by the formula $a_n = a + (n – 1) d$.

For example, to find the general term (or $n^{th}$) term of the progression $3, 11, 19, 27, 35,$ â€¦ we substitute the first term, $a = 3$, and the common difference, $d = 8$ in the formula for the $n^{th}$ term formula.

Using the formula we get, $a_n = a + (n – 1) d = 3 + (n – 1) \times 8 = 3 + 8n – 8 = 8n – 5$. Thus, the general term (or) $n^{th}$ term of this AP is $a_n = 8n – 5$.

### Examples on General Term of Arithmetic Progression

**Example 1:** Find the general term of an AP whose first term is $-5$ and the common difference is $4$.

The first term $a = -5$ and the common difference $d = 4$.

Using the formula $a_n = a + (n – 1)d$, we get $a_n = -5 + (n – 1) \times 4 = -5 + 4n – 4 = 4n – 9$.

**Example 2:** Find the general term of an AP whose first term is $11$ and whose common difference is $-7$.

The first term $a = 11$ and the common difference $d = -7$.

Using the formula $a_n = a + (n – 1)d$, we get $a_n = 11 + (n – 1) \times (-7) = 11 – 7n + 7 = 18 – 7n$.

**Example 3:** Find the general term of an AP $-5, -1, 3, 7, $…

The first term $a = -5$ and the common difference $d = -1 – (-5) = 3 – (-1) = 7 – 3 = â€¦ = 4$.

Using the formula $a_n = a + (n – 1)d$, we get $a_n = -5 + (n – 1) \times 4 = -5 + 4n – 4 = 4n – 9$.

## Derivation of the nth Term of Arithmetic Progression

Let $a_1$, $a_2$, $a_3$, . . . be an AP whose first term $a_1$ is $a$ and the common difference is $d$.

Then, the second term $a_2 = a + d = a + (2 â€“ 1) d$

The third term $a_3 = a_2 + d = (a + d) + d = a + 2d = a + (3 â€“ 1) d$

The fourth term $a_4 = a_3 + d = (a + 2d) + d = a + 3d = a + (4 â€“ 1) d$

Looking at the pattern, we can say that the $n^{th}$ term $a_n = a + (n â€“ 1) d$.

So, the $n^{th}$ term an of the AP with first term $a$ and common difference $d$ is given by $a_n = a + (n â€“ 1) d$.

### Examples on the nth Term of Arithmetic Progression

**Example 1:** Find the $56^{th}$ term of an AP $12, 19, 26, 33, 40, â€¦$.

The first term $a = 12$

The common difference $d = 19 – 12 = 7$

$n = 56$

Formula for $n^{th}$ term an of the AP is $a_n = a + (n â€“ 1) d$.

Substituting the values we get $a_{56} = 12 + (56 â€“ 1) \times 7 = 12 + 55 \times 7 = 397$

Therefore, the $56^{th}$ term of an AP $12, 19, 26, 33, 40, â€¦$ is $397$.

**Example 2: **Which term of the AP $21, 18, 15,$ . . . is $-81$?

The first term $a = 21$

The common difference $d = 18 – 21 = -3$

$a_n = -81$

Formula for $n^{th}$ term an of the AP is $a_n = a + (n â€“ 1) d$.

Substituting the values we get $-81 = 21 + (n â€“ 1) \times (-3) => -81 – 21 = -3n + 3 $

$=> -102 = -3n + 3 => -102 – 3 = -3n => 3n = 105 => n = \frac{105}{3} = 35$.

Therefore the $35^{th}$ term of the AP $21, 18, 15,$ . . . is $-81$.

**Example 3:** Find the $11^{th}$ term from the last term (towards the first term) of the AP $10, 7, 4, . . ., -62$.

The given AP is $10, 7, 4, . . ., -62$, where the first term $a = 10$, and the common difference $d = 7 – 10 = -3$.

Since we want to find the $11^{th}$ term from the last term, letâ€™s reverse the AP.

Now the AP becomes $-62, -59, -56, â€¦ 4, 7, 10$ and the first term $a = -62$, and the common difference $d = -59 – (-62) = 3$ and $n = 11$.

$a_{11} = a + (11 – 1)d = -62 + (11 – 1) \times 3 = -62 + 10 \times 3 = -32$.

Therefore the $11^{th}$ term from the last term (towards the first term) of the AP $10, 7, 4, . . ., -62$ is $-32$.

**Example 4:** Find the $31^{st}$ term of an AP whose $11^{th}$ term is $38$ and the $16^{th}$ term is $73$.

Let the first term and common difference of the AP be $a$ and $d$ respectively.

The $11^{th}$ term of AP is $38$

$=> a + (11 – 1)d = 38 => a + 10d = 38$ â€”———————— (1)

The $16^{th}$ term of AP is $73$

$=> a + (16 – 1)d = 73 => a + 15d = 73$ â€”———————— (2)

Solving (1) and (2), we get $a = -32$ and $d = 7$

Therefore $31^{st}$ term is $a + (31 – 1)d = -32 + 30 \times 7 = -32 + 210 = 178$.

**Example 5:** Determine the AP whose third term is $16$ and the $7^{th}$ term exceeds the $5^{th}$ term by $12$.

Let the first term and common difference of the AP be $a$ and $d$ respectively.

The third term of AP is $a + (3 – 1)d = a + 2d = 16$ â€”——————— (1)

The $7^{th}$ term exceeds the $5^{th}$ term by $12$.

$=> (a + (7 – 1)d) – (a + (5 – 1)d) = 12$

$=> (a +6d) – (a + 4d) = 12 => a +6d – a – 4d = 12 => 2d = 12 => d = 6$

Substituting $d = 6$ in (1), we get $a + 2 \times 6 = 16 => a + 12 = 16 => a = 4$

Therefore the AP is $4, 10, 16, 22, 28, â€¦$

## Sum of n Terms of an Arithmetic Progression

The formula to find the sum of the first $n$ terms of an AP is $\text{S}_n = \frac{n}{2}(2a + (n – 1)d)$, where $a$ is the first term, $d$ is a common difference and $n$ is the total number of terms.

For example, to find the sum of upto $50^{th}$ term of the progression $3, 11, 19, 27, 35,$ â€¦ we substitute the first term, $a = 3$, the common difference, $d = 8$ and $n = 50$ in the sum formula..

Using the formula we get, $\text{S}_{50} = \frac{50}{2}(2 \times 3 + (50 – 1) \times 8) = 25 \times (6 + 49 \times 8) = 25 \times (6 + 392) = 25 \times 398 = 9950$

If the first term and the last term of an AP are known along with the total number of terms of an AP, then the sum of the first $n$ terms of an AP is given by $\text{S}_n = \frac{n}{2}(a + l)$, where $a$ is the first term and $l$ is the last term of an AP.

## Derivation of Sum of n Terms of an Arithmetic Progression

Letâ€™s consider an AP, whose first term is $a$ and the common difference is $d$, then the terms of AP are $a$, $a + d$, $a + 2d$, $a + 3d$, â€¦, $a + (n – 1)d$.

Therefore the sum of the $n$ terms of an AP will be

$\text{S}_n = a + (a + d) + (a + 2d) + (a + 3d) + â€¦ + (a + (n – 1)d)$ â€”———————– (1)

Reversing the terms in (1), we get

$\text{S}_n = (a + (n – 1)d) + (a + (n – 2)d) + (a + (n – 3)d) + (a + (n – 4)d) + â€¦ + a$ â€”———————– (2)

Adding (1) and (2)

$2\text{S}_n = [a + (a + (n – 1)d)] + [(a + d) + (a + (n – 2)d)] + [(a + 2d) + (a + (n – 3)d)] + â€¦ [(a + (n – 1)d) + a]$

$=> 2\text{S}_n = (2a + (n – 1)d) + (2a + (n – 1)d) + (2a + (n – 1)d) + â€¦ n \text{times}$

$=> 2\text{S}_n = n(2a + (n – 1)d)$

$=>\text{S}_n = \frac{n}{2}(2a + (n – 1)d)$ which is the formula for the sum of first $n$ terms of an AP.

$\text{S}_n = \frac{n}{2}(2a + (n – 1)d)$ can be rewritten as $\text{S}_n = \frac{n}{2}(a + (a + (n – 1)d)) = \frac{n}{2}(a + l)$, where $l$ the last term.

### Examples on Sum of n Terms of an Arithmetic Progression

**Example 1:** Find the sum of the first $22$ terms of the AP $8, 3, â€“2, . . $.

The first term and the common difference of the given AP is $a = 8$ and $d = 3 – 8 = -5$ and since we want to find the sum of the first $22$ terms, therefore $n = 22$.

The formula for the sum of the first $n$ terms of an AP is $\text{S}_n = \frac{n}{2}(2a + (n – 1)d)$

Substituting the respective values, we get $\text{S}_{22} = \frac{22}{2}(2 \times 8 + (22 – 1) \times (-5)) = 11 \times (16 + 21 \times (-5)) = 11 \times (16 – 105) 11 \times (-89) = -979$.

Therefore, the sum of the first $22$ terms of the AP $8, 3, â€“2, . . $ is $-979$.

**Example 2:** If the sum of the first $14$ terms of an AP is $1050$ and its first term is $10$, find the $20^{th}$ term.

The first term of an AP $a = 10$

Let the common difference of an AP be $d$.

The sum of the first $14$ terms of an AP is $1050$

$=> \frac{14}{2}(2 \times 10 + (14 – 1)d) = 1050 => 7 \times (20 + 13d) = 1050 => 20 + 13d = \frac{1050}{7}$

$ => 20 + 13d = 150 => 13d = 130 => d = 10$

Now the $20^{th}$ term is $a + (20 – 1)d = 10 + 19 \times 10 = 10 + 190 = 200$.

## Key Takeaways

- An AP is a list of numbers in which each term is obtained by adding a fixed number to the preceding number.
- The first term of AP is represented by $a$, the common difference by $d$, the $n^{th}$ term by $a_n$, and the total number of terms by $n$.
- In general, AP can be represented as $a$, $a + d$, $a + 2d$, $a + 3d$, …
- The $n^{th}$ term of an AP can be obtained by $a_n = a + (n âˆ’ 1)d$.
- The sum of an AP can be obtained by $\text{S}_n= \frac{n}{2} (2a + (n âˆ’ 1) d)$.
- The graph of an AP is a straight line with the slope as the common difference.
- The common difference doesn’t need to be positive always. It can be negative also.

## Practice Problems

- What is meant by Arithmetic Progression?
- Check whether the following sequence of numbers forms an arithmetic progression.
- $4,10,16,22,…$
- $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, â€¦$
- $\sqrt{2}, \sqrt{8}, \sqrt{18}, â€¦$
- $-1.2, -3.2, -5.2, -7.2, …$
- $0.2, 0.22, 0.222, 0.2222, …$

- Find the general term of the following APs
- $2, 9, 16, 23, â€¦$
- $15, 11, 7, 3, -1, â€¦$
- $1, \frac{3}{2}, 2, \frac{5}{2}, â€¦$

- An AP consists of $50$ terms of which $3^{rd}$ term is $12$ and the last term is $106$. Find the $29^{th}$ term.
- If the $3^{rd}$ and the $9^{th}$ terms of an AP are $4$ and $-8$ respectively, which term of this AP is zero?
- The $17^{th}$ term of an AP exceeds its $10^{th}$ term by $7$. Find the common difference.
- Find the sum of the following APs
- $2, 7, 12, . . .$, to $10$ terms
- $â€“37, â€“33, â€“29, . . .$, to $12$ terms
- $0.6, 1.7, 2.8, . . .$, to $100$ terms

- The first term of an AP is $5$, the last term is $45$ and the sum is $400$. Find the number of terms and the common difference.
- The first and the last terms of an AP are $17$ and $350$ respectively. If the common difference is $9$, how many terms are there, and what is their sum?

## FAQs

### What is the meaning of Arithmetic Progression in maths?

A sequence of numbers where each term (other than the first term) is obtained by adding a fixed number to its previous term is called an arithmetic progression (A.P.).Â

For example, $1, 6, 11, 16, 21, â€¦$ is an A.P., where the first term $a = 6$ and the common difference $d = 5$.

### How to find the sum of arithmetic progression?

To find the sum of arithmetic progression, we have to know the first term $a$, the number of terms $n$, and the common difference $d$ between consecutive terms.Â

Then substitute the values in the formula $S_n = \frac{n}{2}(2a + (n âˆ’ 1) d)$.

### How to find the common difference in arithmetic progression?

The common difference is the difference between every two consecutive terms in an arithmetic progression. Therefore, you can say that the formula to find the common difference of an arithmetic sequence is: $d = a_n – a_{n – 1}$, where $a_n$ is the $n^{th}$ term in the progression, and $a_{n – 1}$ is the previous term.

### What is the difference between arithmetic series and arithmetic progression?

Arithmetic progression is a progression in which the difference between every two consecutive terms is a constant whereas an arithmetic series is the sum of the elements of arithmetic progression.

### What is infinite arithmetic progression?

When the number of terms in an AP is infinite, we call it an infinite arithmetic progression. For example, $1, 3, 5, 7, 9, â€¦$ is an infinite AP. The sum of an infinite arithmetic progression doesn’t exist.

## Conclusion

An arithmetic progression abbreviated as AP is a sequence of numbers where the difference between every two consecutive terms is the same. In an AP, each term after the first term is obtained by adding a fixed number generally called a common difference to its previous term. We can find the application of AP in many real-world problems such as determining the number of audience members a stadium can hold or when we take a taxi, we will be charged an initial cost and then a per mile or kilometer charge.