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In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant

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- 1. Arithmetic Progression
- 2. 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.
- 3. • 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
- 4. Common Difference an an 1 4
- 5. 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
- 6. The first term = a1 =a +0 d = a + (11)d second term = a = a + d = a + (2-1)d The 2 The third term = a3 = a + 2d = a + (31)d fourth term = a = a + 3d = a + (4-1)d The 4 ------------------------------------------The nth term = an = a + (n------------------------------------------1)d
- 7. 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.
- 8. (i) 2, 6, 10, 14…. (i) 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.
- 9. . 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
- 10. . 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) Find the AM between 13 & 19 AM= ½(13+ 19) = 16
- 11. Some convenient methods to determine the AP • It is always convenient to make a choice of i. 3 numbers in an AP as (a-d), a, (a+d); ii. 4 numbers in an AP as (a-3d), (a-d), a, (a+d),(a+3d); iii. 5 numbers in an AP as (a-2d), (a-d), a, (a+d), (a+2d).
- 12. 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
- 13. 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 + a n ]
- 14. 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
- 15. Practice 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
- 16. • Solution. 1) First term is an = 500 a = 100 , 9) n - 1 = 80 10) n = 80 + 1 n = 81 2) Common difference is d 11) = 105 -100 = 5 Hence the no. of terms are 3) nth term is an = a + (n-81. 1)d 4) 500 = 100 + (n-1)5 5) 500 - 100 = 5(n – 1) 6) 400 = 5(n – 1) 7) 5(n – 1) = 400 8) n – 1 = 400/5
- 17. 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.
- 18. General Formulas of AP • The general forms of an AP is a,(a+d), (a+2d),. .. , a + ( m - 1)d. i. Nth term of the AP is Tn =a+(n-1)d. ii. Nth term form the end ={l-(n-1)d}, where l is the last term of the word. iii. Sum of 1st n term of an AP is Sn=N/2{2a=(n-1)d}. iv. Also Sn=n/2 (a+l)
- 19. MADE BY-: ,Anirudh Prasad, Ayush Gupta, Chhavi Bansal& Teena Kaushik X- D ROLL NO-6,10 , 12 & 33

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