Maths Notes (Class 10)

CBSE Class 10 Maths Chapter 5 Arithmetic Progressions

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CBSE Class 10 Maths Chapter 5 Arithmetic Progressions
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Arithmetic Progressions

 

Arithmetic Progressions

Arithmetic Progressions

Arithmetic Progression

An arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. For instance, sequence 1, 3, 5, 7, 9, 11, 13, 15 … is an arithmetic progression with common difference of 2. Each of the numbers in the list is called a term.
Real life applications
  1. If you put 10 Rs every day on your piggy bank, you want to know the money after some x days
  2. If salary of a person is Rs 5000 & it increase by 1000 Rs every year, what will be salary after x years?
  3. The lengths of the rungs of a ladder decrease uniformly by 3 cm from bottom to top. If the bottom rung is 50 cm in length, how to find length of any rung?
An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term. E.g.: 2,4,6,8,10 …. This fixed number is called the common difference of the AP. This common difference can be positive, negative or zero
Positive common difference (3): 1,4,7,10,13..
Negative common difference (-1) : 7,6,5,4,3,2,1,0,-1 …
Zero Common Difference : 3,3,3,3,3,3,3,3,3,3….
Let us denote the first term of an AP by a1, second term by a2, . . ., nth term by an and the common difference by d. Then the AP becomes a1, a2, a3, . . ., an.
So, a2 – a1 = a3 – a2 = . . . = an – a– 1 = d.
Thus a, a + d, a + 2d, a + 3d, . . . represents an arithmetic progression where a is the first term and d the common difference. This is called the general form of an AP.
AP with a finite number of terms are called a finite AP E.g.: {1,3,5,7}, while AP with infinite number of terms is called infinite AP, E.g.:{1,3,5,7,…}.
To know about an AP, the minimum information that we need is
  1. First term, denoted by ‘a’
  2. Common difference, denoted by ‘d’
With this we can form the AP as a, a + d, a + 2d, a + 3d, ……
For instance if the first term a is 5 and the common difference d is 2, then the AP is 5, 7,9, 11, . . .
Also, given a list, we can tell if it is AP or not.
List 2,4,6,8,10… here 4-2 = 6-4 = 8-6 = 10-8 =2, thus it is AP with common difference 2
List 1,3,4,6,7, here 3-1 ≠ 4-3, thus it is not AP.
 

nth term of Arithmetic Progression

Let a1, a2, a3, . . . be an AP whose first term a1 is a and the common difference is d., thus the AP is Thus a, a + d, a + 2d, a + 3d,
The first term a1 = a + 0*d = a + (1 – 1) d
The second term a2 = a + d = a + (2 – 1) d
The third term a3 = a + 2d = a + (3 – 1) d
The fourth term a4 = a + 3d = a + (4 – 1) d
. . . . . . . .
Looking at the pattern, we can say that the nth term an = a + (n – 1) d.
So, the nth term an of the AP with first term ‘a’ and common difference d is given by an = a + (n – 1) d.
an is also called the general term of the AP
Example 1: Find the 10th term of the AP : 2, 7, 12, . . .
Solution: Here, = 2, = 7 – 2 = 5 and = 10.
We have an + (– 1) d
So, a10 = 2 + (10 – 1) × 5 = 2 + 45 = 47
Therefore, the 10th term of the given AP is 47.
Example 2: Determine the AP whose 3rd term is 5 and the 7th term is 9.
Solution: a+ (3 – 1) + 2= 5    –(i)
a7 = + (7 – 1) + 6= 9                       –(ii)
Solving the pair of linear equations (i) and (ii), we get a = 3, = 1
Hence, the required AP is 3, 4, 5, 6, 7 . . .
Example 3 : Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . . .
Solution:  da2– a­ = 6 & a=5
Let 301 be a term, say, the nth term of this AP.
an + (– 1) d
or 301 = 5 + (– 1) × 6
301 = 6– 1
Or   n = 302/6 which is a fraction. Thus 301 is not in the series.
 

Sum of first n terms of Arithmetic Progression

Suppose a person gets a salary of 1000 Rs & an increment of 100Rs every month. We want to know how much salary the person has drawn in 10 months. For this we need to find sum of first 10 terms of this AP with a=1000 & d=100.
sum of the first n terms of an AP is given by S = n/2 [2a + (n – 1) d]
We can also write this as S = n/2 [a + a + (n – 1) d]    = n/2 (a + an )
In the case above, n=10, a=1000, d= 100,
So, S = 10/2 [2*1000 + (10-1)*100]    = 5*[2000+900] = 14500.

sum of first positive integers is given by Sn = n(n +1)/2

Also Read :

MATHS Revision Notes

Chapter:01  Real Numbers System
Chapter:02  Polynomials
Chapter:03  Pair of Linear Equations in Two Variables
Chapter:04  Quadratic Equation
Chapter:05  Arithemetic Progressions
Chapter:06  Triangles
Chapter:07  Coordinate Geometry
Chapter:08  Introduction to Trignometry
Chapter:09  Some Application Of Trignometry
Chapter:10  Circles
Chapter:11  Constructions
Chapter:12  Area Related to Cirles
Chapter:13  Surface Area Volume
Chapter:14  Stastistics
Chapter:15  Probability

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