An **arithmetic progression** (AP), or an arithmetic sequence, is a sequence of numbers where the difference between any two consecutive terms in the entire sequence is always constant. This constant term is called the common difference. Let a be the first term and d be the common difference of an AP; then, the arithmetic progression can be written as:

**a, a + d, a + 2d, a + 3d, a + 4d , . . . . . . . . a + (n – 1) d** (Where, n = Total number of terms in an AP)

The nth term of an AP can be written as t_{n} = a + (n – 1)d, where d = a_{n} – a_{n-1}

The sum of 1st n terms of an arithmetic progression is given by:

**S _{n} = (n/2) **

**[2a + (n – 1) d]**

The last term of an arithmetic progression is **l = a + (n – 1) d**. Therefore, **S _{n} = (n/2)**

**[a + l]**

**Properties of Arithmetic Progression (AP)**

(1) The value of the common difference (d) of an AP is either a positive or negative number. The AP will be an **increasing arithmetic sequence,** if the value of the common difference (d) is positive. Similarly, the AP will be a **decreasing arithmetic sequence,** if the value of the common difference (d) is positive.

**Example:** 12, 24, 36, 48, . . . . . . . , 120 is an increasing sequence with initial term (a) = 12, common difference (d) = 12 and an = 120.

Similarly, 24, 12, 0, – 12, – 24, . . . . . . – 120 a decreasing sequence with the initial term (a) = 24, common difference (d) = – 12 and an = – 120.

In this case, the common difference (D) is negative, i.e., (d < 0); hence, a_{1} > a_{2} > a_{3} > . . . . . . > a_{n}.

(2) If each term of an arithmetic progression (AP) is decreased, increased, divided or multiplied by the same non-zero number, then the resulting sequence will also form an AP.

**Example:** Consider an arithmetic progression 4, 8, 12, 16, . . . . . Here, the initial term, a = 4 and the common difference, d = 4. If we add every term of the given sequence by 4, then the resulting sequence 8, 12, 16, 20, . . . will also form an AP with a = 8 and d = 12.

Similarly, if we multiply every term of the given sequence by 2, then the resulting sequence 8, 16, 24, 32, . . . . . will also form an AP with a = 8 and d = 8.

(3) If m, n, and o are 3 consecutive terms of an AP, then 2n = m + o. Also, if m, n, o, and p are 4 consecutive terms of an arithmetic progression, then m + p = n + o.

(4) In an arithmetic progression, the sum of any 2 terms equidistant from the beginning and end is always constant. Also, their sum is always equal to the sum of the 1st term and the last term. Let a_{1}, a_{2}, a_{3}, a_{4},…, a_{n-1}, a_{n} forms an AP, then,

**\(\begin{array}{l}a_{2} + a_{n – 2} = a_{3} + a_{n – 3} = a_{4} + a_{n – 4} = a_{1}\ (\text{First term})+a_{n}\ (\text{Last term})\end{array} \)**

(5) Any term of an arithmetic progression (excluding the initial term) is equal to half the sum of terms which are equidistant from it. Let a_{1}, a_{2}, a_{3}, a_{4},…, a_{n-1}, a_{n} forms an AP, then,

(6) If the nth term of any sequence is a linear expression in n, i.e., Pn + Q, then the sequence is an AP, and the common difference of that arithmetic progression is P.

(7) If the sum of n terms of any sequence is quadratic in n, i.e.,

**Middle Term of an Arithmetic Progression**

If there are n number of terms in an AP, then,

**Case 1: **If n is odd.

The middle term of an arithmetic progression is given by** **

**\(\begin{array}{l}\mathbf{\left ( \frac{n\;+\;1}{2} \right )^{th}}\end{array} \)**

**Case 2:** If n is even.

Middle terms of an arithmetic progression are given by

**\(\begin{array}{l}\mathbf{\left ( \frac{n}{2} \right )^{th}}\ \text{and}\ \mathbf{\left ( \frac{n}{2}\;+\;1 \right )^{th}}\end{array} \)**

**Arithmetic Progression – Important Tips**

- The sum of the first n natural numbers is given by

**\(\begin{array}{l}\mathbf{\sum \;n\;=\;\frac{n\;(n\;+\;1)}{2},\;n\;\in \;N}\end{array} \)**

- The sum of the squares of the first n natural numbers is given by

**\(\begin{array}{l}\mathbf{\sum \;n^{2}\;=\;\frac{n\;(n\;+\;1)\;(n\;+\;2)}{6},\;n\;\in \;N}\end{array} \)**

- The sum of the cubes of the first n natural numbers is given by

**\(\begin{array}{l}\mathbf{\sum \;n^{3}\;=\;\left [ \frac{n\;(n\;+\;1)}{2} \right ],\;n\;\in \;N}\end{array} \)**

- Three conceive terms of an AP can be taken as: (a – d), a, and (a + d). Similarly, four consecutive terms of an AP can be taken as (a – 2d), (a – d), (a + d), and (a + 2d).

- Formation of an arithmetic progression.

An arithmetic progression can be formed by inserting two or more numbers between any two given numbers. Consider P_{1}, P_{2}, P_{3}, P_{4}, P_{5},…., P_{n} be n numbers between k and m, such that the resulting series k, P_{1}, P_{2}, P_{3}, P_{4}, P_{5},…., P_{n}, m forms an AP. The resulting arithmetic progression will have a total of (n + 2) terms.

Hence, m = a + (n – 1) d = k + [(n + 2) – 1] d = k + (n + 1) d [Since, m is (n + 2)th term]

Now,

Similarly

Therefore, for nth term,

Using this method, n numbers can be inserted between two numbers, such that the resulting sequence is an AP.

**Example: Insert 4 numbers between 12 and 72, such that the resulting sequence is an arithmetic progression.**

**Solution:**

Let P_{1}, P_{2}, P_{3}, and P_{4} be the four numbers between 12 and 72, such that the sequence 12, P_{1}, P_{2}, P_{3}, P_{4}, 72 forms an AP.

Here, k (Initial term) = 12, m (Last term) = 72, n (Total no. of terms) = 6. Therefore, 72 = 12 + (6 -1) d

i.e., 60 = 5d or d = 12

Hence, P_{1} = a + d = 12 + 12 = 24, P_{2} = a + 2d = 12 + 24 = 36, P_{3} = a + 3d = 12 + 36 = 48, P_{4} = a + 4d = 12 + 48 = 60.

Hence, the resulting sequence is 12, 24, 36, 48, 60, and 72.

**Arithmetic Progressions IIT JEE Problems**

**Example 1: **

**\(\begin{array}{l}\text{If}\ S_{n}=\sum_{1}^{4\;n}\;(-\;1)^{\frac{k\;(k\;+\;1)}{2}}\;k^{2}.\end{array} \)**

**Check if S**

_{n}**can take the following values: 1056 and 1088.**

**Solution:**

Given:

i.e.

Or,

Or,

Or,

Or,

Since,

Therefore,

i.e., S_{n} = n [8 + 8 n – 8 + 12 + 8n – 8] = 4n ( 4n + 1)

Now, 1088 can be expressed as 32 × 34 = 4.8 (4.8 + 2)

Hence, the value of Sn cannot be 1088.

And, 1056 can be expressed as 32 × 33 = 4.8 (4.8 + 1)

Hence, the value of Sn can be 1056.

**Example 2: Find the value of the tenth term of an AP, if the 4th term and 54th term of an arithmetic progression are 64 and – 61, respectively.**

**Solution:**

Given, T_{4} = 64 = a + (4 – 1) d and T_{54} = – 61 = a + (54 – 1) d

i.e., a + 3d = 64 . . . . . . . . . (1)

And, a + 53d = – 61 . . . . . . . . (2)

On subtracting equation (1) and equation (2), we will get

50 d = – 125 or d = -5/2

On substituting the values of d in equation (1), we will get a = 143/2

Now, T_{10} = a + 9d

**\(\begin{array}{l}=\frac{143}{2} + 9\times (-\frac{5}{2})\end{array} \)**

Therefore, T_{10} = 49

**Example 3: Find the 50th**** term of a sequence if the sum of n terms is given by** S_{n} = 2n^{2} + 3n.

**Solution:**

Given:

The nth term of the sequence is given by

i.e.,

Or,

Hence, T_{50} (50th Term) = 4 (50) + 1 = 201.

**Example 4: Find the values of a, b, c, and d if,**

**\(\begin{array}{l}\mathbf{\sum_{k\;=\;1}^{n}\;\left [ \;\sum_{m\;=\;1}^{k}\; m^{2\;}\right ]\;=\;an^{4}\;+\;bn^{3}\;+\;cn^{2}\;+\;dn\;+\;e}\end{array} \)**

**Solution:**

Since the sum of the squares of first n natural numbers is given by:

**\(\begin{array}{l}\mathbf{\sum \;n^{2}\;=\;\frac{n\;(n\;+\;1)\;(n\;+\;2)}{6},\;n\;\in \;N}\end{array} \)**

Therefore,

**\(\begin{array}{l}\mathbf{\sum_{k\;=\;1}^{n}\;\left [ \;\sum_{m\;=\;1}^{k}\; m^{2\;}\right ]\;=\;\sum_{k\;=\;1}^{n}\;\frac{k\;(k\;+\;1)\;(2k\;+\;1)}{6}}\end{array} \)**

**\(\begin{array}{l}=\mathbf{\frac{1}{6}\;\sum_{k\;=\;1}^{n}\;2k^{3}\;+\;3k^{2}\;+\;k}\end{array} \)**

Since the sum of the cubes of first n natural numbers is given by:

**\(\begin{array}{l}\mathbf{\sum \;n^{3}\;=\;\left [ \frac{n\;(n\;+\;1)}{2} \right ],\;n\;\in \;N}\end{array} \)**

Therefore,

**\(\begin{array}{l}\mathbf{\frac{1}{3}\;\left [ \frac{n\;(n\;+\;1)}{2} \right ]^{2}\;+\;\frac{1}{2}\;\left [ \frac{n\;(n\;+\;1)\;(2n\;+\;1)}{6} \right ]\;+\;\frac{1}{12}\;n\;(n\;+\;1)}\end{array} \)**

**\(\begin{array}{l}=\mathbf{\frac{1}{12}\;\left [n^{4}\;+\;n^{2}\;+\;2n^{3}\right ]\;+\;\frac{1}{12}\;\left [2n^{3}\;+\;3n^{2}\;+\;n\right ]\;+\;\frac{n^{2}}{12}\;+\;\frac{n}{12}}\end{array} \)**

**\(\begin{array}{l}=\frac{1}{6}n^{4}+\frac{1}{6}n^{3}+\frac{n^{2}}{3}+\frac{5}{12}n+\frac{1}{6}\end{array} \)**

**\(\begin{array}{l}=\mathbf{\frac{1}{12}\;n^{4}\;+\;\frac{1}{3}\;n^{3}\;+\;\frac{5}{12}\;n^{2}\;+\;\frac{1}{6}\;n}\end{array} \)**

On comparing the coefficients of a, b, c, d, and e from the above equation, we get

a = 1/12, b = 1/3, c = 5/12, d = 1/6 and e = 0.

**Example 5:** The pth term of the series

**Solution:**

The given series is

Therefore common difference

Now,

Trick: This question can be solved by inspection.

**Example 6: **Which of the following sequences is an arithmetic sequence?

**Solution: **

Sequence f (n) = an + b; n ∈ N is an A.P.

Putting n = 1, 2, 3, 4, ………., we get the sequence

(a + b), (2a + b), (3a + b),……… which is an A.P. where first term (A) = (a + b) and common difference d = a.

Aliter: As we have mentioned in the theory part, the nth term of an A.P. is of the form an + b, n ∈ N.

**Formation of an Arithmetic Progression**

An arithmetic progression can be formed by inserting two or more numbers between any two given numbers. Consider P_{1}, P_{2}, P_{3}, P_{4}, P_{5},…., P_{n} be n numbers between k and m, such that the resulting series k, P_{1}, P_{2}, P_{3}, P_{4}, P_{5},…., P_{n}, m forms an AP. The resulting arithmetic progression will have a total of (n + 2) terms.

Hence, m = a + (n – 1) d = k + [(n + 2) – 1] d = k + (n + 1) d [Since, m is (n + 2)th term]

i.e.,

Now,

Similarly,

Therefore, for the nth term,

Using this method, n numbers can be inserted between two numbers, such that the resulting sequence is an AP.

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## Frequently Asked Questions

### What do you mean by arithmetic progression?

An arithmetic progression (AP) is a sequence of numbers where the difference between each consecutive term is the same. For example, 3, 5, 7, 9. Here, 2 is the common difference.

### What do you mean by common difference in AP?

The common difference is the difference between two consecutive terms in an arithmetic sequence. It is denoted by d. d = a_{2} – a_{1} or a_{n} – a_{n-1}.

### Give the formula for the sum of n terms of AP.

The sum of n terms of AP is given by S_{n} = (n/2)(2a+(n-1)d). Here, a is the first term of AP, and d is the common difference. If the first term and last term of AP are given, then S_{n} = (n/2)(first term + last term).

### Give the formula for the n^{th} term of A.P.

The n^{th} term of an A.P is given by t_{n} = a+(n-1)d, where a is the first term and d is the common difference.

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