Arithmetic Progression SUBMITTED BY- PRIYARANJAN KUMAR CLASS-10 ‘A’ ROLL NO.-17

Arithmetic Sequence Arithmetic Sequence is a sequence of numbers such that the difference between the consecutive terms is constant. For instance, the sequence 5, 7, 9, 11, 13, 15 … is an arithmetic progression with common difference of 2. 2,6,18,54(next term to the term is to be obtained by multiplying by 3. Arithmetic Sequence

Arithmetic series A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of a finite arithmetic progression is called an Arithmetic series. The behavior of the arithmetic progression depends on the common difference d. If the common difference is: Positive, the members (terms) will grow towards positive infinity. Negative, the members (terms) will grow towards negative infinity.

Common Difference If we take first term of an AP as a and Common Difference as d. Then-- nth term of that AP will be An = a + (n-1)d. For instance--- 3, 7, 11, 15, 19 … d =4 a =3 Notice in this sequence that if we find the difference between any term and the term before it we always get 4. 4 is then called the common difference and is denoted with the letter d. To get to the next term in the sequence we would add 4 so a recursive formula for this sequence is: The first term in the sequence would be a1 which is sometimes just written as a.

Arithmetic Progression If the initial term of an arithmetic progression is a1 and the common difference of successive members is d, then the nth term of the sequence (an) is given by: An= a1+(n-1)d And in general An= a1+ (n-m)d

To find the nth Term of an A.P. Let us consider an A.P. with first term ‘a’ and common difference ‘d’ ,then 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 The nth term = an = a + (n-1)d -------------------------------------------

Example 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.

To check that a given term is in A.P. or not. 2, 6, 10, 14…. Here first term a = 2, find differences in the next terms a2-a1 = 6 – 2 = 4 a3-a2 = 10 –6 = 4 a4-a3 = 14 – 10 = 4 Since the differences are common. Hence the given terms are in A.P.

Problem 1. Find 10th term of A.P. 12, 18, 24, 30.. … Solution. Given A.P. is 12, 18, 24, 30.. First term is a = 12 Common difference is d = 18- 12 = 6 nth term is an = a + (n-1)d Put n = 10, a10 = 12 + (10-1)6 = 12 + 9 x 6 = 12 + 54 a10 = 66

Arithmetic Mean Practice Let A be the AM between a and b. To find the Arithmetic Mean Between Two Numbers. Let A be the AM between a and b. A, A, b are in AP (A-a)=(b-A) A=1/2(a+b) AM between a and b A=1/2(a+b) Practice Find the AM between 13 & 19 AM= ½(13+ 19) = 16

Some convenient methods to determine the AP It is always convenient to make a choice of 3 numbers in an AP as (a-d), a, (a+d); 4 numbers in an AP as (a-3d), (a-d), a, (a+d),(a+3d); 5 numbers in an AP as (a-2d), (a-d), a, (a+d), (a+2d).

Practice Q---- The sum of three numbers in an AP is 21 & their product is 231. Find the numbers. Let the required Numbers be (a-d), a, (a+d) Then (a-d)+ a+ (a+d)=21 A=7 And (a-d) a (a+d)=231 A(a2-d2) =231 Substituting the value of ‘a’ 7(49-d2)=231 7d2= (343-231)= 112 D2= 16 D= 4

Sum of n-term of an ap Sn = ½ n [ a + an ] The sum of n terms, we find as, Sum = n X [(first term + last term) / 2] Now last term will be = a + (n-1) d Therefore, Sn = ½ n [ 2a + (n - 1)d ] It can also be written as Sn = ½ n [ a + an ]

The sum to n terms of an arithmetic progression Problem 1. 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 = ½ n [ 2a + (n - 1)d ] = ½ x 30 [ 2x 12 + (30-1)x 8] = 15 [ 24 + 29 x8] = 15[24 + 232] = 15 x 246 = 3690

Practice Q3: Given a = 2, d = 8, Sn = 90, find n and an. Q1: Find the sum of the following APs: (i) 2, 7, 12, ……, to 10 terms Q3: Given a = 2, d = 8, Sn = 90, find n and an. Question: 2. Find the sums given below: (i) 7+10.5+14+…..+84

Problem-- Find number of terms of A.P. 100, 105, 110, 115……500 Solution. n - 1 = 80 First term is a = 100 , an = 500 n = 80 + 1 Common difference is d = 105 -100 = 5 n = 81 Hence the no. of terms are --81. nth term is an = a + (n-1)d 500 = 100 + (n-1)5 500 - 100 = 5(n – 1) 400 = 5(n – 1) 5(n – 1) = 400 n – 1 = 400/5

Practice Question: 1. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149. Question: 2. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

General Formulas of AP The general forms of an AP is a,(a+d), (a+2d),. .. , a + ( m - 1)d. Nth term of the AP is Tn =a+(n-1)d. Nth term form the end ={l-(n-1)d}, where l is the last term of the word. Sum of 1st n term of an AP is Sn=N/2{2a=(n-1)d}. Also Sn=n/2 (a+l)

THANK YOU