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

**\(\begin{array}{l}\mathbf{ t_{n} = a + (n – 1) d }\end{array} \)**

**,**where

The Sum of 1st n terms of an Arithmetic Progression is given by:

**\(\begin{array}{l}S_{n}\end{array} \)**

**=**

**\(\begin{array}{l}\mathbf{\frac{n}{2}}\end{array} \) **

**[2a + (n – 1) d]**

Since, the last term of an Arithmetic Progression is, **l = a + (n – 1) d**. Therefore,

**Properties of Arithmetic Progression (AP)**

1.The value of the common difference (d) of an AP is either 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, a1 > a2 > a3 > a3 > . . . . . . > an.

- If each term of an Arithmetic Progression (AP) is decreased, increased, divided or multiplied or 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.

- 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.
- 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 1st term and the last term. Let, \(\begin{array}{l}a_{1}, a_{2}, a_{3}, a_{4}, . . . a_{n – 1}, a_{n}\end{array} \)forms an AP, then,

**\(\begin{array}{l}a_{2} + a_{n – 2}\end{array} \)**

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

- Any term of an Arithmetic Progression (excluding the Initial term) is equal to the half the sum of terms which are equidistant from it. Let, \(\begin{array}{l}a_{1}, a_{2}, a_{3}, a_{4}, . . . a_{n – 1}, a_{n}\end{array} \)forms an AP, then,

- 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.
- If the sum of n terms of any sequence is quadratic in n i.e. \(\begin{array}{l}Pn^{2} + Qn + R,\end{array} \)then the sequence is an AP and the common difference of that Arithmetic Progression is 2P.

**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:

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

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

**Arithmetic Progression Important tips:**

- The Sum of 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 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 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

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

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,

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

**Arithmetic Progressions IIT JEE Problems:**

**Example 1: If **

**\(\begin{array}{l}S_{n}\end{array} \)**

**=**

**\(\begin{array}{l}\mathbf{\sum_{1}^{4\;n}\;(-\;1)^{\frac{k\;(k\;+\;1)}{2}}\;k^{2}}\end{array} \)**

**. Check if\(\begin{array}{l}S_{n}\end{array} \)**

**can take the following values: 1056 and 1088.**

**Solution:**

Given: If

i.e.

Or,

Or,

Or,

Or,

Since,

Therefore,

i.e.

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 tenth term of an AP if, 4th term and 54th term of an arithmetic progression are 64 and – 61 respectively.**

**Solution:**

Given,

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 =

**\(\begin{array}{l}\mathbf{-\;\frac{5}{2}}\end{array} \)**

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

**\(\begin{array}{l}\mathbf{\frac{143}{2}}\end{array} \)**

Now,

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

Therefore,

**Example 3: Find 50th**** term of a sequence if the sum of n terms is given by**

**Solution:**

Given:

The nth term of the sequence is given by

i.e.

Or,

Hence,

**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 =

**\(\begin{array}{l}\mathbf{\frac{1}{12}}\end{array} \)**

**, d =\(\begin{array}{l}\mathbf{\frac{5}{12}}\end{array} \)**

**and e = 0.\(\begin{array}{l}\mathbf{\frac{1}{6}}\end{array} \)**

**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 theory part that 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

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

i.e. d =

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

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,

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

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## Frequently Asked Questions

### What do you mean by arithmetic progression?

An arithmetic progression (A.P) 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 A.P?

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 A.P.

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

### Give the formula for 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.