Arithmetic Progression Class 10 ||Maths|| Chapter 5 NCERT Notes

Arithmetic Progression Class 10 ||Maths|| Chapter 5 NCERT Notes

Arithmetic Progression Class 10 ||Maths|| Chapter 5 NCERT Notes

1. What is an Arithmetic Progression (AP)?

An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference (d).

Definition of Arithmetic Progression:

A sequence of numbers is said to be in AP if the difference between any two consecutive terms is constant.
If $a_{1}, a_{2}, a_{3}, \dots$ are the terms of an AP, then the common difference $d = a_{2} - a_{1} = a_{3} - a_{2} = \dots$ .

Example of an AP:

$2, 5, 8, 11, 14, \dots$ Here, the common difference $d = 5 - 2 = 3$ .

General form of an AP:

The terms of an AP are generally represented as: $a, a + d, a + 2 d, a + 3 d, \dots$
Where:
$a$ is the first term.
$d$ is the common difference.

2. Important Terms and Formulae in AP

a. Common Difference (d):

The common difference in an AP is the difference between any two consecutive terms.
Formula:
$d = a_{n + 1} - a_{n}$ Where $a_{n + 1}$ is the next term, and $a_{n}$ is the current term.

b. The nth Term of an AP (General Term):

The nth term of an arithmetic progression gives the value of any term located at the nth position in the sequence.
Formula:
$a_{n} = a + (n - 1) \cdot d$
Where:
$a_{n}$ is the nth term.
$a$ is the first term.
$n$ is the position of the term.
$d$ is the common difference.

Example:

Find the 10th term of the AP: $3, 7, 11, \dots$ .
Here, $a = 3$ , $d = 7 - 3 = 4$ , and $n = 10$ .
Using the formula:
$a_{n} = a + (n - 1) \cdot d = 3 + (10 - 1) \cdot 4 = 3 + 36 = 39$
So, the 10th term is 39.

3. Sum of the First n Terms of an AP (S_n)

The sum of the first n terms of an AP is the total sum obtained by adding the first n terms of the sequence.

Formula for the Sum of the First n Terms:

$S_{n} = \frac{n}{2} [2 a + (n - 1) \cdot d]$
Where:
$S_{n}$ is the sum of the first $n$ terms.
$a$ is the first term.
$d$ is the common difference.
$n$ is the number of terms.
Alternatively, if the first term $a$ and last term $l$ are known, we can use:
$S_{n} = \frac{n}{2} (a + l)$
Where $l$ is the nth term.

Example:

Find the sum of the first 15 terms of the AP: $4, 9, 14, 19, \dots$ .
Here:
$a = 4$
$d = 9 - 4 = 5$
$n = 15$
Using the sum formula:
$S_{15} = \frac{15}{2} [2 \times 4 + (15 - 1) \times 5] = \frac{15}{2} [8 + 70] = \frac{15}{2} \times 78 = 15 \times 39 = 585$
So, the sum of the first 15 terms is 585.

4. Special Cases in AP

a. If the Common Difference is Zero:

If the common difference $d = 0$ , all the terms in the sequence are the same.
Example: $5, 5, 5, 5, 5, \dots$

b. AP with Negative Common Difference:

When the common difference $d$ is negative, the terms decrease as the sequence progresses. Example: $10, 7, 4, 1, - 2, \dots$ where $d = - 3$ .

5. Problems Related to AP

a. Finding the Number of Terms (n):

Sometimes, you're given the first term, the last term, and the common difference, and you need to find how many terms are in the AP.
Formula:
Using the nth term formula:
$a_{n} = a + (n - 1) \cdot d$
Solve for $n$ .

Example:

Find how many terms are there in the AP: $7, 13, 19, \dots, 205$ .
Here:
$a = 7$
$d = 13 - 7 = 6$
$a_{n} = 205$
Using the formula for the nth term:
$205 = 7 + (n - 1) \cdot 6$ $205 - 7 = (n - 1) \cdot 6$ $198 = (n - 1) \cdot 6$ $n - 1 = \frac{198}{6} = 33$ $n = 33 + 1 = 34$
So, there are 34 terms in the AP.

b. Inserting Arithmetic Means:

If you need to insert k arithmetic means between two numbers $a$ and $b$ , the total number of terms in the sequence becomes $k + 2$ , with the two given numbers as the first and last terms, respectively.
The common difference is calculated as:
$d = \frac{b - a}{k + 1}$

Example:

Insert 3 arithmetic means between 10 and 34.
Here:
$a = 10$
$b = 34$
$k = 3$
The common difference:
$d = \frac{34 - 10}{3 + 1} = \frac{24}{4} = 6$
The arithmetic progression will be:
$10, 16, 22, 28, 34$
So, the 3 arithmetic means are 16, 22, and 28.

6. Word Problems Based on AP

APs are used in a variety of real-life situations, and the chapter includes word problems to demonstrate this. Here’s an example of how AP can be applied in practical scenarios:

Example:

The sum of the first 10 terms of an AP is 150. If the first term is 7, find the common difference.
Given:
$S_{10} = 150$
$a = 7$
$n = 10$
Using the sum formula:
$S_{n} = \frac{n}{2} [2 a + (n - 1) \cdot d]$
Substitute the values:
$150 = \frac{10}{2} [2 \times 7 + (10 - 1) \cdot d]$ $150 = 5 [14 + 9 d]$ $150 = 70 + 45 d$ $150 - 70 = 45 d$ $80 = 45 d$ $d = \frac{80}{45} = \frac{16}{9}$
So, the common difference $d = \frac{16}{9}$ .

7. Applications of Arithmetic Progression

AP is not just a theoretical concept. It has practical applications in everyday life, such as:
Calculating installments in finance.
Predicting trends or patterns in economics, population growth, or scientific data.
Organizing seating arrangements or numbering systems.
This completes the detailed explanation of Arithmetic Progressions for Class 10 Maths. If you need further clarification or more examples, feel free to ask!