# Arithmetic Progression – Common difference and Nth term | Class 10 Maths

Arithmetic Progression is a sequence of numbers where the difference between any two successive numbers is constant. For example

1, 3, 5, 7, 9……. is in a series which has a common difference (3 – 1) between two successive terms is equal to 2. If we take natural numbers as an example of series 1, 2, 3, 4… then the common difference (2 – 1) between the two successive terms is equal to 1.

In other words, arithmetic progression can be defined as “**A mathematical sequence in which the difference between two consecutive terms is always a constant**“.

We come across the different words like sequence, series, and progression in AP, now let us see what does each word define –

Sequenceis a finite or infinite list of numbers that follows a certain pattern. For example 0, 1, 2, 3, 4, 5â€¦ is the sequence, which is infinite sequence of whole numbers.Seriesis the sum of the elements in which the sequence is corresponding For example 1 + 2 + 3 + 4 + 5â€¦.is the series of natural numbers. Each number in a sequence or a series is called a term. Here 1 is a term, 2 is a term, 3 is a term …….Progressionis a sequence in which the general term can be expressed using a mathematical formula or the Sequence which uses a mathematical formula that can be defined as the progression.

## What is the Common difference of an AP?

The common difference in the arithmetic progression is denoted by** d**. The difference between the successive term and its preceding term. It is always constant or the same for arithmetic progression. In other words, we can say that, in a given sequence if the common difference is constant or the same then we can say that the given sequence is in Arithmetic Progression.

- The formula to find common difference is
**d = (a**._{n + 1 }– a_{n }) or d = (a_{n}– a_{n-1}) - If the common difference is positive, then AP increases. For Example 4, 8, 12, 16….. in these series, AP increases
- If the common difference is negative then AP decreases. For Example -4, -6, -8……., here AP decreases.
- If the common difference is zero then AP will be constant. For Example 1, 2, 3, 4, 5………, here AP is constant.

The sequence of Arithmetic Progression will be like a_{1}, a_{2,} a_{3}, a_{4},…

**Example 1: 0, 5, 10, 15, 20…..**

here, a_{1 }= 0, a_{2}= 5, so a_{2}- a_{1}= d = 5 - 0 = 5. a_{3}= 10, a_{2 }= 5, so a_{3}- a_{2 }= 10 - 5 = 5. a_{4 }= 15, a_{3}= 10, so a_{4 }- a_{3 }= 15 - 10 =5. a_{5 }=20, a_{4 }=15, so a_{5 }-a_{4 }= 20 - 15 = 5.

From the above example, we can say that the common difference is “5”.

**Example 2: 0, 7, 14, 21, 28…….**

here, a_{1}= 0, a_{2}= 7, so a_{2}- a_{1 }= 7 - 0 = 7 a_{3}= 14, a_{2}= 7, so a_{3}- a_{2}= 14 - 7 = 7 a_{4}= 21, a_{3}= 14, so a_{4}- a_{3}= 21 - 14 = 7 a_{5}=28, a_{4}= 21, so a_{5 }-a_{4}=28 - 21 = 7

From the above example, we can say that the common difference is “7”.

**How to Find the Middle term of an AP?**

To find the middle term of an arithmetic progression we need the total number of terms in a sequence. We have two cases:

Even:If the number of terms in the sequence is even then we will be having two middle terms i.e (n/2) and (n/2 + 1).n

Odd:If the number of terms in the sequence is odd then we will be having only one middle terms i.e (n/2).

**Example 1:**

If n = 9 then, Middle term = n/2 = 9/2 = 4.

**Example 2:**

If n = 16 then, First middle term = n/2 = 16/2 = 8. Second middle term = (n/2) + 1 = (16/2) + 1 = 8 + 1 = 9.

**What is the Nth term of an AP?**

To find the nth term of an arithmetic progression, We know that the A.P series is in the form of a, a + d, a + 2d, a + 3d, a + 4d……….

The n^{th } term is denoted by **T _{n. }**Thus to find the nth term of an A.P series will be :

**Example: Find the 9 ^{th} term of the given A.P sequence: 3, 6, 9, 12, 15………..?**

**Step 1: **Write the given series.

Given series = 3, 6, 9, 12, 15...........

**Step 2**: Now write down the value of a and n from the given series.

a = 3, n = 9

**Step 3: **Find the common difference d by using the formula **(a _{n+1 } – a_{n})**.

d = a2 - a1 , here a2 = 6 and a1 = 3 so d = (6 - 3) = 3.

**Step 4**: We need to substitute values of a, d, n in the formula (T_{n }= a + (n – 1)d).

T_{n}= a + (n - 1)d given n = 9. T_{9}= 3 + (9 - 1)3 = 3 + (8)3 = 3 + 24 = 27

Therefore the 9^{th } term of given A.P series 3, 6, 9, 12, 15………. is “27”.