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 **\(\mathbf{ t_{n} = a + (n – 1) d }\), **where \(d = a_{n} – a_{n -1}\)

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

**\(S_{n}\) = ****\(\mathbf{\frac{n}{2}}\) ****[2a + (n – 1) d]**

Since, the last term of an Arithmetic Progression is, **l = a + (n – 1) d**. Therefore, \(S_{n}\) = \(\mathbf{\frac{n}{2}}\) [a + l]

**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, \(a_{1}, a_{2}, a_{3}, a_{4}, . . . a_{n – 1}, a_{n}\) forms an AP, then,

**\(a_{2} + a_{n – 2}\) = \(a_{3} + a_{n – 3} = a_{4} + a_{n – 4} = a_{1}\) (First Term) + \(a_{n}\) (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, \(a_{1}, a_{2}, a_{3}, a_{4}, . . . a_{n – 1}, a_{n}\) forms an AP, then,

\(a_{5}\) = \(\mathbf{\frac{1}{2}}\) [latex]a_{1} + a_{9}[/latex]= \(\mathbf{\frac{1}{2}}\) \([a_{1} + a_{9}]\) = \(\mathbf{\frac{1}{2}}\) \([a_{2} + a_{8}]\) = \(\mathbf{\frac{1}{2}}\) \([a_{3} + a_{7}]\) and so on.

- 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. \(Pn^{2} + Qn + R,\) 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** \(\mathbf{\left ( \frac{n\;+\;1}{2} \right )^{th}}\)**

**Case 2:** If n is Even

Middle Terms of an Arithmetic Progression are given by:

**\(\mathbf{\left ( \frac{n}{2} \right )^{th}}\)** and** \(\mathbf{\left ( \frac{n}{2}\;+\;1 \right )^{th}}\)**

**Arithmetic Progression Important tips:**

- The Sum of first n natural numbers is given by:

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

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

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

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

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

- 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]

\(d = \frac{m-k}{n+1}\)Now, \( P_{1} = a + d = k + \frac {m − k} {n + 1} \) [Since, n = 1 (Second Term)]

Similarly \( P_{2} = a + 2d = k + 2 \frac {m − k} {n + 1} \) [Since, n = 1 (Third Term)]

\( P_{3} = a + 3d = k + 3 \frac {m − k} {n + 1} \) [Since, n = 1 (Fourth Term)]

\( P_{4} = a + 4d = k + 4 \frac {m − k} {n + 1} \) [Since, n = 1 (Fifth Term)]

Therefore, for nth term, \( P_{n} = a + nd = k + n\frac {m − k} {n + 1} \)

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, \(P_{1}\), \(P_{2}\), \(P_{3}\), and \(P_{4}\), 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, 72.

**Arithmetic Progressions IIT JEE Problems:**

**Example 1: If \(S_{n}\) = ****\(\mathbf{\sum_{1}^{4\;n}\;(-\;1)^{\frac{k\;(k\;+\;1)}{2}}\;k^{2}}\)****. Check if \(S_{n}\)** **can take the following values: 1056 and 1088.**

**Solution:**

Given: If \(S_{n}\) = \(\mathbf{\sum_{1}^{4\;n}\;(-\;1)^{\frac{k\;(k\;+\;1)}{2}}\;k^{2}}\)

i.e. \(S_{n}\) = \(– 1^{2} – 2^{2} + 3^{2} + 4^{2} – 5^{2} – 6^{2} + 7^{2} + 8^{2} – 9^{2} – 10^{2} + 11^{2} + 12^{2} +\) . . . . . . up to 4n terms.

Or, \(S_{n} = (3^{2} – 1^{2}) + (42 – 2^{2}) + (7^{2} – 5^{2}) + (8^{2} – 6^{2}) + (11^{2} – 9^{2}) + (12^{2} – 10^{2}) + \) . . . . . up to 4n terms.

Or, \(S_{n} = 8 + 12 + 24 + 28 + 40 + 44 + \). . . . . . up to 4n terms.

Or, \(S_{n} = 2 (4 + 6 + 12 + 14 + 20 + 22 + \) . . . . . up to 2n terms).

Or, \(S_{n} \) = 2 [(4 + 12 + 20 + . . . up to n terms) (6 + 14 + 22 + . . . up to n terms)].

Since, \(S_{n} \) = \(\mathbf{\frac{n}{2}\;\left [ \;2a\;+\;\left ( n\;-\;1 \right )\;d\; \right ]}\)

Therefore, \(S_{n} \) = \(\mathbf{2\;\left | \;\frac{n}{2} \left [ 2\;\times \;4\;+\;\left ( n\;-\;1 \right )\;8 \right ]\;+\;\frac{n}{2}\;\left [ 2\;\times \;6\;+\;(n\;-\;1)\;8 \right ]\right |}\)

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 tenth term of an AP if, 4****th**** term and 54****th**** 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 = **\(\mathbf{-\;\frac{5}{2}}\)**

On substituting the values of d in equation (1) we will get a = **\(\mathbf{\frac{143}{2}}\)**

Now, \(T_{10}\) = a + 9d = **\(\mathbf{\frac{143}{2}}\) + \(\mathbf{9\;\times \;-\;\frac{5}{2}}\)**

Therefore, \(T_{10}\) = 49

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

**Solution:**

Given: \(S_{n} = 2n^{2} + 3n\)

The nth term of the sequence is given by \(T_{n} = S_{n} – S_{n – 1}\)

i.e. \(T_{n} = 2n^{2} + 3n – [ 2 (n -1)^{2} + 3(n – 1) ] = 2n^{2} + 3n – 2n^{2} – 2 + 4n – 3n + 3\)

Or, \(T_{n} = 4n + 1\)

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

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

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

**Solution:**

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

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

Therefore, **\(\mathbf{\sum_{k\;=\;1}^{n}\;\left [ \;\sum_{m\;=\;1}^{k}\; m^{2\;}\right ]\;=\;\sum_{k\;=\;1}^{n}\;\frac{k\;(k\;+\;1)\;(2k\;+\;1)}{6}}\)**

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

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

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

Therefore, **\(\mathbf{\frac{1}{3}\;\left [ \frac{n\;(n\;+\;1)}{2} \right ]^{2}\;+\;\frac{1}{2}\;\left [ \frac{n\;(n\;+\;1)\;(2n\;+\;1)}{6} \right ]\;+\;\frac{1}{12}\;n\;(n\;+\;1)}\)**

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

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

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

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

a = **\(\mathbf{\frac{1}{12}}\)**, b = **\(\mathbf{\frac{1}{3}}\)**, c = **\(\mathbf{\frac{5}{12}}\)**, d =** \(\mathbf{\frac{1}{6}}\)** and e = 0.

**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. d = \(\mathbf{\frac{m\;-\;k}{n\;+\;1}}\)

Now, \(P_{1}\) = a + d = k + \(\mathbf{\frac{m\;-\;k}{n\;+\;1}}\) [Since, n = 1 (Second Term)]

Similarly \(P_{2}\) = a + 2 d = k + \(\mathbf{\frac{2\;(m\;-\;k)}{n\;+\;1}}\) [Since, n = 1 (Third Term)]

\(P_{3}\) = a + 3 d = k + \(\mathbf{\frac{3\;(m\;-\;k)}{n\;+\;1}}\) [Since, n = 1 (Fourth Term)]

\(P_{4}\) = a + 4 d = k + \(\mathbf{\frac{4\;(m\;-\;k)}{n\;+\;1}}\) [Since, n = 1 (Fifth Term)]

Therefore, for nth term, \(P_{n}\) = a + nd = k + \(\mathbf{\frac{n\;(m\;-\;k)}{n\;+\;1}}\)

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 \(P_{1}, P_{2}, P_{3}, P_{4}\) 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, 72.