- Introduction
- |
- What is arithmetic progression
- |
- Some Important points about AP
- |
- nth term of Arithmetic Progression
- |
- Sum of nth item in Arithmetic Progression

So here we see the pattern in Rohan salary. The difference between two consecutive year salaries is constant. In the similar, world around us present us these pattern in many form. In this chapter, we will learn a special type of pattern called Arithmetic Progression

Let us consider following series

All these sets follow certain rules. In first set $5 - 1 = 9 - 5 = 13 - 9 = 17 - 13 = 4$

In second set $2 - 1 = 3 - 2 = 4 - 3 = 1$

and so on.

Here the difference between any successive members is a constant

Such series are called Arithmetic Progression

- The difference is called the common difference of the AP and It is denoted by d
- The members are called terms. The first member is called first term
- We can denote common difference by d
- If $a_1, a_2,a_3,a_4,a_5$ are the terms in AP then

$D=a_2 - a_1 =a_3 - a_2 =a_4 - a_3=a_5 - a_4$ - We can represent the general form of AP in the form

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

Where $a$ is first term and $d$ is the common difference - If the AP series has last term then it s finite Arithmetic Progression and if the AP series has infinite then it is called the Inifinite Arithmetic Progression

$1, 4, 7, 10, 13, 16, 19, 22, 25,...$

$D=3=3=3=3$

So it is AP

Observe the number sequence and complete the next two terms of the sequence

- $2, 8, 14, 20$
- $5, 13, 21, 29$
- $11, 22, 33, 44$
- $-3, 1,5, 9$
- $5, -1, -7, -13$

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

Here $a = 1$, $d = 4$

$a + d = 1 + 4 = 5$

$a + 2d = 1 + 2 x 4 = 9$

and so on

Thus we can say that

$a$ = First term

$a + d$ = Second term

$a + 2d$ = Third term

$a + 3d$ = Fourth term and son on

$n_{th} \; term = a + (n - 1)d$

Here First term = $t_1= a$

Second term = $t_2 = a + d$

and hence, $t_n= a + (n - 1)d$

d is called common difference and the series is called arithmetic progression

Find the 13th term of the AP : 1, 5, 9, . . .

Here, $a = 1, d = 5 - 1 = 4 \; and \; n = 13$

We have $a_n = a + (n - 1) d$

So, $a_{13} = 1 + (13 - 1) \times 5 = 1 + 60 = 61$

Therefore, the 13th term of the given AP is 61.

Which term of the AP : 21, 18, 15, . . . is -81? Also, is any term 0? Give reason for your answer.

Here, $a = 21, d = 18 - 21 = - 3 \; and \; a_n = -81$, and we have to find n.

As $a_n = a + (n - 1) d$

we have $-81 = 21 + (n - 1)( -3)$

$- 81 = 24 - 3n$

$-105 = -3n$

So, n = 35

Therefore, the 35th term of the given AP is -81.

Next, we want to know if there is any n for which $a_n = 0$. If such an n is there, then

$21 + (n - 1) (-3) = 0$

i.e., $3(n - 1) = 21$

i.e., n = 8

So, the eighth term is 0.

Determine the AP whose 3rd term is 5 and the 7th term is 9.

We have

$a_3 = a + (3 - 1) d = a + 2d = 5$ (1)

and $a_7 = a + (7 - 1) d = a + 6d = 9$ (2)

Solving the pair of linear equations (1) and (2), we get

a = 3, d = 1

Hence, the required AP is 3, 4, 5, 6, 7, . . .

$S = 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1$

Adding these two we get $2s = 11 + 11 + 11 + 11 + 11 + 11 + 11 = 11 + 11 + 11$

$2S= 10 \times 11$

$S = \frac {(10 \times 11)}{2} = 55$

In similar way, if

Adding we get

Find the sum of first 24 terms of the list of numbers whose nth term is given by

$a_n = 5 + 2n$

As $a_n = 5 + 2n$

so, $a_1 = 5 + 2 = 7$

$a_2 = 5 + 2 \times 2 = 9$

$a_3 = 5 + 2 \times 3 = 11$

So,

List of numbers becomes 7, 9, 11, . . .

Here, 9 - 7 = 11 - 9 = 2 and so on.

So, it forms an AP with common difference d = 2.

To find $S_{24}$, we have n = 24, a = 7, d = 2.

Therefore, $S_{24} = \frac {24}{2}[2 \times 7+(24 -1) 2]$ = 12 [14 + 46] = 720

So, sum of first 24 terms of the list of numbers is 720

Given below are the links of some of the reference books for class 10 math.

- Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
- Mathematics for Class 10 by R D Sharma
- Pearson IIT Foundation Maths Class 10
- Secondary School Mathematics for Class 10
- Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.

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