Arithmetic-Progressions bvgc cfdfd cg-4.pptx

Arithmetic Progression (AP) is a sequence of numbers in order, in which the difference between any two consecutive numbers is a constant value. It is also called Arithmetic Sequence. For example, the series of natural numbers: 1, 2, 3, 4, 5, 6,… is an Arithmetic Progression, which has a common difference between two successive terms (say 1 and 2) equal to 1 (2 -1). Even in the case of odd numbers and even numbers, we can see the common difference between two successive terms will be equal to 2.

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If we observe in our regular lives, we come across Arithmetic progression quite often. For example, Roll numbers of students in a class, days in a week or months in a year. This pattern of series and sequences has been generalized in Maths as progressions.

Table of Contents
Definition
Notation
First Term
Common Difference
General Form
Nth Term
Types of AP
Sum of Nth Term
Formula List
Questions and Solutions
Problems to Solve
FAQs

What is Arithmetic Progression?
In mathematics, there are three different types of progressions. They are:

Arithmetic Progression (AP)
Geometric Progression (GP)
Harmonic Progression (HP)
A progression is a special type of sequence for which it is possible to obtain a formula for the nth term. The Arithmetic Progression is the most commonly used sequence in maths with easy to understand formulas.

Definition 1: A mathematical sequence in which the difference between two consecutive terms is always a constant and it is abbreviated as AP.

Definition 2: An arithmetic sequence or progression is defined as a sequence of numbers in which for every pair of consecutive terms, the second number is obtained by adding a fixed number to the first one.

The fixed number that must be added to any term of an AP to get the next term is known as the common difference of the AP. Now, let us consider the sequence, 1, 4, 7, 10, 13, 16,…

It is considered as an arithmetic sequence (progression) with a common difference 3. Notation in Arithmetic Progression
In AP, we will come across some main terms, which are denoted as:

First term (a)
Common difference (d)
nth Term (an)
Sum of the first n terms (Sn)
All three terms represent the property of Arithmetic Progression. We will learn more about these three properties in the next section.

First Term of AP
The AP can also be written in terms of common differences, as follows;

a, a + d, a + 2d, a + 3d, a + 4d, ………. ,a + (n – 1) d
where “a” is the first term of the progression.

Common Difference in Arithmetic Progression
In this progression, for a given series, the terms used are the first term, the common difference and nth term. Suppose, a1, a2, a3, ……………., an is an AP, then; the common difference “ d ” can be obtained as;

d = a2 – a1 = a3 – a2 = ……. = an – an – 1
Where “d” is a common difference. It can be positive, negative or zero.

1. ARITHMETIC PROGRESSION

2. If various terms of a sequence are formed by adding a fixed number to the previous term or the difference between two successive terms is a fixed number, then the sequence is called AP. For example : 5, 10, 15, 20, 25….. In this each term is obtained by adding 5 to the preceding term except first term

3. The general form of an Arithmetic Progression is a , a +d , a + 2d , a + 3d ………………, a + (n-1)d Where, ‘a’ is first term and ‘d’ is called common difference.

4. Common Difference - The fixed number which is obtained by subtracting any term of AP from its succeeding term. If we take first term of an AP as a and Common Difference as d , Then, n th term of that AP will be a n = a + (n-1)d

5. To check that a given term is in A.P. or not. 2, 6, 10, 14…. Here first term a = 2, Now, find differences in the next terms a 2 -a 1 = 6 – 2 = 4 a 3 -a 2 = 10 –6 = 4 a 4 -a 3 = 14 – 10 = 4 Since the differences are common. Hence the given terms are in A.P.

6. Problem : Find the value of k for which the given series is in A.P. 4, k –1 , 12 Solution : Given A.P. is 4, k –1 , 12….. If series is A.P. then the differences will be common. a 2 – a 1 = a 3 – a 2  k – 1 – 4 = 12 – (k – 1)  k – 5 = 12 – k + 1  k + k = 12 + 1 + 5  2 k = 18 or k = 9

7. Its formula is SUM OF n TERMS OF AN ARITHMETIC PROGRESSION It can also be written as S n = ½ n [ 2a + (n - 1)d ] S n = ½ n [ a + a n ]

8. DERIVATION The sum to n terms is given by: S n = a + (a + d) + (a + 2d) + … + [a + (n – 1)d]     (1) If we write this out backwards, we get: S n = [a + (n – 1)d] + (a + (n – 2)d) + … +a            (2) Now let’s add (1) and (2): 2S n = [2a + (n – 1)d] + [2a + (n – 1)d] + … ……… + [2a + (n – 1)d] So, S n = ½ n [2a + (n – 1)d]

9. Problem . Find number of terms of A.P. 100, 105, 110, 115,,………………500 Solution. First term is a = 100 , a n = 500 Common difference is d = 105 -100 = 5 nth term is a n = a + (n-1)d  500 = 100 + (n-1)5  500 - 100 = 5(n – 1)  400 = 5(n – 1)  5(n – 1) = 400

10.  5(n – 1) = 400  n – 1 = 400/5  n - 1 = 80  n = 80 + 1  n = 81 Hence the no. of terms in the AP are 81.

11. Problem Find the sum of 30 terms of given A.P. ,12 , 20 , 28 , 36……… Solution : Given A.P. is 12 , 20, 28 , 36 Its first term is a = 12 Common difference is d = 20 – 12 = 8 The sum to n terms of an AP S n = ½ n [ 2a + (n - 1)d ] = ½ x 30 [ 2 x 12 + (30-1)x 8]

12. = 15 [ 24 + 29 x 8] = 15[24 + 232] = 15 x 246 = 3690 THE SUM OF TERMS IS 3690

13. Problem . Find the sum of terms in given A.P. 2 , 4 , 6 , 8 , ……………… 200 Solution: Its first term is a = 2 Common difference is d = 4 – 2 = 2 nth term is a n = a + (n-1)d  200 = 2 + (n-1)2  200 - 2 = 2(n – 1)  2(n – 1) = 198  n – 1 = 99, n = 100

14. Now, the sum to n terms of an arithmetic progression S n = ½ n [ 2a + (n - 1)d ] S 100 = ½ x 100 [ 2x 2 + (100-1)x 2] = 50 [ 4 + 198] = 50[202] = 10100