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We see lot of pattern in our daily lives. For example, Rohan salary in his first year is 2000 and he will be getting increment of 500 every year,So his salary for First, second, third, fourth …. Will be of the form 2000,2500,3000,3500
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
What is arithmetic progression?
An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant
Let us consider following series (1)$1,5,9,13,17....$ (2)$1,2,3,4,5,...$ (3)$ 7,7,7,7.....$
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
Some Important points about AP
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
Example:
Find if the below series is Arithmetic Progression
$1, 4, 7, 10, 13, 16, 19, 22, 25,...$ Solution:
$D=3=3=3=3$
So it is AP Practice Questions
Observe the number sequence and complete the next two terms of the sequence
$2, 8, 14, 20$
$5, 13, 21, 29$
$11, 22, 33, 44$
$-3, 1,5, 9$
$5, -1, -7, -13$
nth term of Arithmetic Progression
On the basis of above discussion we can consider the following series
$a, a + d, a + 2d, a + 3d, .........................$
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
Example 1
Find the 13th term of the AP : 1, 5, 9, . . . Solution
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.
Example 2
Which term of the AP : 21, 18, 15, . . . is -81? Also, is any term 0? Give reason for your answer. Solution
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. Example 3
Determine the AP whose 3rd term is 5 and the 7th term is 9. Solution
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, . . .
Sum of n items in Arithmetic Progression
Let $S = 1 + 2 + 3 + 4 + 5 + 6 + 7 + + 8 + 9 + 10$ and writting in reversed order
$S = 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1$
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
Example
Find the sum of first 24 terms of the list of numbers whose nth term is given by
$a_n = 5 + 2n$ Solution
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
Arithmetic Mean
If a,b,c are in AP, then $b=\frac {(a+c)}{2}$ and b is called the Arithmetic Mean
Practice Questions
Quiz Time
Summary
Here is the Arithmetic Progressions Class 10 Maths Notes Summary
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
nth term is given by $a_n =a + (n-1)d$
Sum of n term $S_n= \frac {n}{2} [2a + (n-1)d]$
Arithmetic mean is give by $A= \frac {a+b}{2}$
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