CLASSX MATHEMATICS ARITHMETIC PROGRESSION Learning Objectives Students will

CLASS-X MATHEMATICS ARITHMETIC PROGRESSION

Learning Objectives Students will be able to 1. recall some patterns which occur in their day to day life. 2. know the condition when the pattern will be in arithmetic progression. 3. know various terms associated with AP, i. e 1 st term and common difference. 4. understand different types of AP, finite & infinite. 5. understand the formula to find nth term and sum of n terms of an AP. 6. apply the knowledge of finding nth term and sum of n terms of an AP in solving day to day

Sequence �A sequence is a set of numbers written in a particular order �Examples of sequence: Ø 1, 3, 5, 7, 9, . . Ø 1, 4, 9, 16, 25, . . Ø 1, − 1, . . . , Ø 1, 3, 5, 9

Series �A series is something we obtain from a sequence by adding all the terms together. �For example, let us consider the sequence of numbers 1, 2, 3, 4, 5, 6, . . . , n. � Then S 1 = 1, as it is the sum of just the first term on its own. �The sum of the first two terms is S 2 = 1 + 2 = 3. � Continuing, we get S 3 = 1 + 2 + 3 = 6 ,

Arithmetic progressions �An arithmetic progression, or AP, is a sequence where each new term after the first is obtained by adding a constant d, called the common difference, to the preceding term. �If the first term of the sequence is a then the arithmetic progression is a, a + d, a + 2 d, a + 3 d, . . . Where d is the common difference

To check that a given terms are in A. P. or not. 2, 6, 10, 14…. Here first term a =2, 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 difference between the consecutive terms is same Hence the given terms are in A. P.

EXAMPLE 1 : 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. d 1 =d 1 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 K=9

ACTIVITY-1 �Objective: To verify that the given sequence is an arithmetic progression by paper cutting and pasting method. �Materials required: coloured paper, pair of scissors, geometry box, fevicol, sketch pens, one squared paper

ACTIVITY-1(continued. . ) �Procedure : Take a given sequence of numbers say a 1 , a 2 , a 3 …. 2. Cut a rectangular strip from a colored paper of width k = 1 cm (say) and length a 1 cm. 3. Repeat this procedure by cutting rectangular strips of the same width k = 1 cm and lengths a 2 , a 3 , a 4 , … cm. 4. Take 1 cm squared paper and paste the rectangular strips adjacent to each other in order.

ACTIVITY-1(continued. . ) �Let the sequence be 1, 4, 7, 10, …. Take strips of lengths 1 cm, 4 cm, 7 cm and 10 cm, all of the same width say 1 cm. Arrange the strips in order as shown in Fig 2(a). Observe that the adjoining strips have a common difference in heights. (In this example it is 3 cm. )

ACTIVITY-1(continued. . )

ACTIVITY-1(continued. . ) �Let another sequence be 1, 4, 6, 9, … Take strips of lengths 1 cm, 4 cm, 6 cm and 9 cm all of the same width say 1 cm. Arrange them in an order as shown in Fig 2(b). Observe that in this case the adjoining strips do not have the same difference in heights.

ACTIVITY-1(continued. . )

Conclusion �So, from the figures, it is observed that if the given sequence is an arithmetic progression, a ladder is formed in which the difference between the heights of adjoining steps is constant. If the sequence is not an arithmetic progression, a ladder is formed in which the difference between adjoining steps is not constant.

nth Term of an AP �a, a + d, a + 2 d, a + 3 d, a + 4 d, . . . is an AP. where a is the first term, and d is the common difference. � If we wanted to write down the n-th term, we would have a + (n − 1)d ,

Let’s see an example EXAMPLE 1: Let a=2, d=2, n=12, find An An=a+(n-1)d =2+(12 -1)2 =2+(11)2 =2+22 Therefore, An=24 Hence solved.

nth Term of an AP Example-2: Write down the first five terms of the AP with first term 8 and common difference 7. Ans: 8, 15, 22, 29, 36 Example-3: Write down the first five terms of the AP with first term 2 and common difference − 5. Ans: 2, -3, -8, -13, -18

nth Term of an AP Example 4: What is the common difference of the AP 11, − 13, − 25, . . . ? Ans: common difference = -1 -11 =-13 -(-1) =-12

nth Term of an AP Example-5: Find the 17 th term of the arithmetic progression with first term 5 and common difference 2. Ans:

The sum of an arithmetic series �Sometimes we want to add the terms of a sequence. Let us see the video https: //youtu. be/S 6 F 6 je. VX-b 8 �What would we get if we wanted to add the first n terms of an arithmetic progression? �We would get Sn = a + (a + d) + (a + 2 d) +. . . + (ℓ − 2 d) + (ℓ − d) + ℓ. �This is now a series, as we have added together the n terms of a sequence. This is an arithmetic series, and we can find its sum by using a trick.

The sum of an arithmetic series �Let us write the series down again, but this time we shall write it down with the terms in reverse order. �We get Sn = ℓ + (ℓ − d) + (ℓ − 2 d) +. . . + (a + 2 d) + (a + d) + a � 2 Sn = (a + ℓ) +. . . + (a + ℓ), � 2 Sn = n(a + ℓ)

The sum of an arithmetic series �Sn = ½ n(a + ℓ). �ℓ = a + (n − 1)d , �Sn = ½ n(a + ℓ) �Sn = ½ n(a + (n − 1)d) � Sn= ½ n(2 a + (n − 1)d).

The sum of an arithmetic series �Example -1: Find the sum of the first 50 terms of the sequence 1, 3, 5, 7, 9, . . Solution : This is an arithmetic progression, a = 1 , d = 2 , n = 50. Sn = ½ n(2 a + (n − 1)d) S 50 = ½ × 50 × (2 × 1 + (50 − 1) × 2) = 25 × (2 + 49 × 2) = 25 × (2 + 98) = 2500

The sum of an arithmetic series �Example-2: Find the sum of the series 1 + 3· 5 + 6 + 8· 5 +. . . + 101. Solution : Given a=1 , d= 3. 5 - 1 = 2. 5 and ℓ=101 we know ℓ = a + (n − 1)d 101 = 1 + (n − 1) × 2· 5. 100 = (n − 1) × 2· 5 40 = n − 1 n = 41 So, Sn = ½ n(a+ ℓ)= ½ × 41(1+101)=2091

The sum of an arithmetic series �Example -3: An arithmetic progression has 3 as its first term. Also, the sum of the first 8 terms is twice the sum of the first 5 terms. Find the common difference. Solution : Given a = 3. S 8= ½ × 8 × (6 + 7 d), S 5= ½ × 5 × (6 + 4 d) S 8 = 2 S 5, we see that ½ × 8 × (6 + 7 d) = 2 × ½ × 5 × (6 + 4 d) 4 × (6 + 7 d) = 5 × (6 + 4 d) 24 + 28 d = 30 + 20 d 8 d = 6

Example 4: 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) =400

5(n n – – 1) 1 = = 400/5 n - 1 = 80 n =80 +1 n =81 Hence the no. of term is 81.

Example 5. 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 arithmetic progression Sn=n/2 [ 2 a +(n - 1)d ] = ½ x 30 [ 2 x 12 + (30 -1)x 8] =15 [ 24 +29 x 8]

=15[24 + 232] 15 x 246 �= �= 3690 �THE SUM OF TERMS IS 3690

Summary ØIn this chapter, you have studied the following points : ØAn arithmetic progression (AP) is a list of numbers in which each term is obtained by adding a fixed number d to the preceding term, except the first term. The fixed number d is called the common difference. ØThe general form of an AP is a, a + d, a + 2 d, a + 3 d, . . .

Summary �A given list of numbers a 1, a 2, a 3, . . . is an AP, if the differences a 2 – a 1, a 3 – a 2, a 4 – a 3, . . . , give the same value, i. e. , if ak + 1 – ak is the same for different values of k. � In an AP with first term a and common difference d, the nth term (or the general term) is given by �an = a + (n – 1) d.

Summary �The sum of the first n terms of an AP is given by : S n= ½ n[2 a+(n-1)d] �If l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP is given by : Sn = ½ n(a+l)

MIND MAP Sum of first n positive integers How many 2 -digit numbers are divisible by 3? 2 -digit numbers divisible by 3 12, 15, 18, . . . 99 a = 12, d = 3, a n = 99 a n = a + (n– 1)d 99 = 12 + (n– 1)3 s n = n ( a + l ) = n(1+n) 2 2 n(n+l) or s n = 2 s n = n (a+ l) 2 87 = 29 3 n = 30 n– 1 = List of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term. Fixed number is called common difference. er Arithmet If a, b, c, are in AP, b= a+c 2 ic Progressi on b is arithmetic mean When first term of common differnce is given : �� �� = {2�� + (�� − 1)�� } �� 2 a, a+d, a+2 d, a+3 d, . . . a+(n – 1) d Fixed number in arithmetic progression which provides the to and fro terms by adding/ subtracting from the present number. h Te i. e. , Let s n = 1 + 2 + 3 +. . . n a = 1, last term l = n From beginning a n = a+(n– 1)d Here a – first term a- First When first & last terms are given: d- common d �� �� = (�� + �� ) – �� �� 2 co m on di difference Can be positive or negative.