Our CBSE Notes for Introduction of Arithmetic Progression Class 10 maths 5. Arithmetic Progressions - CBSE Notes - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Introduction of Arithmetic Progression Class 10 maths 5. Arithmetic Progressions - CBSE Notes - Toppers Study

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## Introduction of Arithmetic Progression Class 10 maths 5. Arithmetic Progressions - CBSE Notes - Toppers Study

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### Introduction of Arithmetic Progression class 10 Mathematics Chapter 5. Arithmetic Progressions

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- Introduction Of Arithmetic Progression Class 10 Maths 5. Arithmetic Progressions - CBSE Notes - Toppers Study
- Class 10 Ncert Solutions
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- Solutions Class 10
- Chapter 5. Arithmetic Progressions Introduction Of Arithmetic Progression Class 10

## 5. Arithmetic Progressions

### | Introduction of Arithmetic Progression |

## Introduction of Arithmetic Progression Class 10 maths 5. Arithmetic Progressions - CBSE Notes - Toppers Study

**Introduction of Arithmetic Progression**

**Sequence:** A set of numbers arranged in some definite order and formed

according to some rules is called a sequence.

**Progression:** The sequence that follows a certain pattern is called

progression.

**Arithmetic Progression: **An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.

**Examples:**

(a) 1, 2, 3, 4, 5 ......................................

(b) 2, 5, 8, 11, 14 ..................................

(c) 50, 45, 40, 35 .....................................

(d) -10, -4, 2, 8 .........................................

In above examples, each of the numbers in the list is called a **term**. Each term is obtained by adding a fixed number except the first term. And we can write other next term by adding same fixed number. This fixed number is called the **common difference** of the AP.

In other words, A list of numbers which has same common difference will be said **Arithmetic progression**.

- Each number of an arithmetic progression is called t
**erm**. - The first number of A.P is called
**first term**. - The last term of A.P is called f
**inal term**. - Difference between two consicutive terms of an A.P is known as
**common difference**.

The first term is denoted by a or a_{1. }and

Second term by a_{2}

Third term by a_{3}

Fourth term by a_{4}

and so on.

Common difference by **d**

the number of term is denoted by **n**

Final term is denoted by a_{n}

**Important points and Formula: **

(A) General form of an A.P is

a, a + d, a + 2d, a + 3d, a + 4d ............................................................

(where a is the first term and d is common difference)

so the,

(B) General term for n^{th} term is **a _{n} = {a + (n -1)d}**

(C) If **a**, **b**, and **c** are in A.P then

This is also Known as Arithmetic Mean.

(D) If three numbers are in A.P then take **a - d, a, a + d** as general term.

(E) General formula for **the sum of natural numbers** begin with 1 like

1 + 2 + 3 .......

(F) Sum of n^{th} of an A. P:

(G) If ‘l’ is the last term of a finite A.P., then the sum is given by

(H) If S_{n} is given, then n^{th} term is given by a_{n} = s_{n} – s_{n -1}.

**General formula for common difference:**

d = a_{n} - a_{n -1}

Like,

d = a_{2 }- a_{1}

d = a_{3} - a_{2}

d = a_{5 }- a_{4}

and so on ...

### 1. Checking for A. P

**Question 1: Is 2, 4, 8, 16, . . . an A.P?**

**Solution:**

d = a_{2} - a_{1} = 4 - 2 = 2

d = a_{3} - a_{1} = 8 - 4 = 4,

d = a_{4} - a_{3} = 16 - 8 = 8,

Here all the value of Common difference (**d)** is not a fixed number or common, these all are different like 2, 4 and 8.

Hence 2, 4, 8, 16 .............. is not an A.P.

**Question2: Is – 10, – 6, – 2, 2, . . . an A. P?**

**Solution:**

d = a_{2} - a_{1} = - 6 - (-10) = - 6 +10 = 4,

d = a_{3} - a_{1} = - 2 - ( -6) = - 2 + 6 = 4,

d = a_{4} - a_{3} = 2 - (-2) = 2 + 2 = 4

Here all the value of common difference (d) is fixed or equal.

Hence – 10, – 6, – 2, 2, . . . is an A.P.

### Writing Next terms of the A.P`

Question 1: Write the next four terms of the A.P, when the first term a and the common difference d are given as follows:

**(i) a = 4, d = -3 ** **(ii) a = 7, d = 13**

**Solution: **

(i) general form of an A.P is

a, a + d, a + 2d, a + 3d ..............

4, 4 + (-3), 4 + 2(-3), 4 + 3(-3) ........

4, 4 - 3, 4 - 6, 4 - 9 .........

4, 1, -2, - 5 ............

Hence required A.P is 4, 1, -2, - 5 ............

(ii) general form of an A.P is

a, a + d, a + 2d, a + 3d ..............

7, 7 + 13, 7 + 2(13), 7 + 3(13) .........

7, 20, 33, 46 ................

Hence required A.P is 7, 20, 33, 46 ................

### Finding First Term And Common difference:

**Question1: For the following APs, write the first term and the common difference:**

**(i) 3, 1, – 1, – 3, . . . ** ** (ii) – 5, – 1, 3, 7, . . .**

**Solution: **

(i) a = 3,

d = a_{n} - a_{n -1 }

_{ }∴_{ }d = a_{2} - a_{1}

_{ }= 1 - 3 = - 2

First term = 3, common difference = - 2

(ii) – 5, – 1, 3, 7, . . .

a = - 5

d = a_{2} - a_{1}

= - 1 - (-5)

= - 1 + 5

= 4

First term = - 5, common difference = 4

### Finding n^{th} term And Number of term:

**Question1: From the given APs, find the n ^{th} term.**

(i) 17, 13, 9, 5 ........................ 33^{th }term

(ii) - 5, 0, 5, 10 ................... 24^{th} term

**Solution: **

**(i) 17, 13, 9, 5 ........................ **

a = 17, d = 13 - 17 = - 4, n = 33

a_{n} = a + (n -1)d

a_{33} = a + (33 - 1)d

= a + 32d

= 17 + 32(-4)

= 17 + (-128)

= 17 - 128

= - 111

_{ }∴ 33^{th} term is - 111.

(ii) - 5, 0, 5, 10 ................... 24^{th} term

a = -5,

d = 0 -(-5) = 0 +5 = 5,

n = 24,

a_{24} = a + 23d

= - 5 + 23 (5)

= - 5 + 115

= 110

**Question2: Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.**

**Solution: **

**Method 1:** Solving from first term.

a = 3, d = 8 - 3 = 5, a_{n} = 253

// Here we have to find first n of final term 253 //

a_{n} = a + (n -1)d

253 = 3 + (n - 1)5

253 - 3 = (n -1)5

250 = (n - 1)5

n - 1 = 250 / 5

n - 1 = 50

n = 50 + 1

n = 51

We have to find 20^{th} term from last term

∴ required n will be 51 - 19 = 32

a_{32} = a + 31d

= 3 + 31(5)

= 3 + 155 = 158

Hence 20^{th }term from last term is 158

**Method 2: **solving from final term.

a = 253, d = 3 - 8 = - 5, n = 20

a_{20 }= a + 19d

= 253 + 19(-5)

= 253 - 95

= 158

Hence 20^{th }term from last term is 158

##### Other Pages of this Chapter: 5. Arithmetic Progressions

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