AP&GP.pptx

aps

1. UNIT-3 ARITHMETIC AND GEOMETRIC PROGRESSIONS

2. Arithmetic Progression (AP) is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value. For example, the series of natural numbers: 1, 2, 3, 4, 5, 6,… is an AP, which has a common difference between two successive terms (say 1 and 2) equal to 1 (2 -1).

3. In AP, we will come across three main terms, which are denoted as: Common difference (d) nth Term (a n ) Sum of the first n terms (S n ) Where “d” is a common difference. It can be positive, negative or zero.

4. FORMULA General Form of AP- a, a + d, a + 2d, a + 3d, . . The nth term of AP - t n  = a + (n – 1) × d Sum of n terms in AP- S n = n/2[2a +(n − 1)× d]

5. CONTS… n th term of A.P. t n  = a + (n – 1) × d a – first terms d – common difference (2 nd term – 1 st term) For eg. A.p. 2,4,6,8…. 4 -2 =2 6 - 2=2 8 - 6=2

6. Where  a = First term d = Common difference n = number of terms a n  = nth term

7. Type 1 Example 1: Find the 20th term for the given AP:3, 5, 7, 9, ……

8. Solution:  Given,  3, 5, 7, 9, …… a = 3, d = 5 – 3 = 2, n = 20 a n = a + (n − 1) × d a 20 = 3 + (20 − 1) × 2 a 20 = 3 + 38 ⇒ a 20 = 41

9. Example 2: Find the 100th term for the given AP:5, 10, 15, ……

10. Solution:  Given,  a = 5, d = 5 , n = 100 a n = a + (n − 1) × d a 100 = 5 + (100 − 1) × 5 ⇒ a 100 = 500

11. Example 3: Find the 17th term of the Ap with first term five and common differences two

12. Solution:  Given,  a = 5, d = 2 , n = 17 a n = a + (n − 1) × d a 17 = 5 + (17 − 1) × 2 ⇒ a 17 = 37

13. Example 4: Find the 10th term of the Ap with first term 8 and common differences 3

14. Solution:  Given,  a = 8, d = 3 , n = 10 a n = a + (n − 1) × d a 10 = 8 + (10 − 1) × 3 ⇒ a 10 = 35

15. Type 2 Example 1: Find the value of n. If a = 10, d = 5, last term= 95.

16. Solution:  Given, a = 10, d = 5, a n = 95 From the formula of general term, we have: a n = a + (n − 1) × d 95 = 10 + (n − 1) × 5 (n − 1) × 5 = 95 – 10 = 85 (n − 1) = 85/ 5 (n − 1) = 17 n = 17 + 1 n = 18

17. Example 2: Find the value of n. If a = 5, d = 5, last term= 500.

18. Solution:  Given, a = 5, d = 5, a n = 500 From the formula of general term, we have: a n = a + (n − 1) × d 500 = 5 + (n − 1) × 5 n = 100

19. Example 3 : Find the value of n. If a = 8, d = 3, last term= 62.

20. Solution:  Given, a = 8, d = 3, a n = 62 From the formula of general term, we have: a n = a + (n − 1) × d 62 = 8 + (n − 1) × 3 n = 19

21. Example 4 : Find the value of n. If a = 8, d = 3, last term= -19.

22. Solution:  Given, a = 8, d = -3, a n = - 19 From the formula of general term, we have: a n = a + (n − 1) × d -19 = 8 + (n − 1) × -3 n = 10

23. Type 3 Sum of N Terms of AP Consider an AP consisting “n” terms. Example 1 : Let us take the example of adding natural numbers up to 15 numbers.

24. AP = 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 Given, a = 1, d = 2-1 = 1 and an = 15 Now, by the formula we know; S = n/2[2a + (n − 1) × d] = 15/2[2x1+(15-1)x1] S = 15/2[2+14] = 15/2 [16] = 15 x 8 S = 120 Hence, the sum of the first 15 natural numbers is 120.

25. Example 2 : Find the sum of the first 50 term of the sequence 1,3,5,7,9…... Given, a = 1, d = 2 and an = 50 Now, by the formula we know; S = n/2[2a + (n − 1) × d] = 50/2[2x1+(50-1)x2] S = 2500

26. Example 3 : Find the sum of the first 23 term of the sequence 4,-3,-10,…...

27. Given, a = 4, d = - 7and an = 23 Now, by the formula we know; S = n/2[2a + (n − 1) × d] = 23/2[2x4+(23-1)x-7] S = - 1679

28. Example 4 : Find the sum of the first 20 term of the sequence, first term four and common differences half.

29. Given, a = 4, d = 1/2 and an = 20 Now, by the formula we know; S = n/2[2a + (n − 1) × d] = 20/2[2x4+(20-1)x1/2] S = 175

30. Type 4 Sum of AP when the Last Term is Given First we need to find out n term Second as per follow concern formula Example 1 : 1+3.5+6+8+5+....+101

31. Question 1: Find 10 th term in the series 1,3,5,7… A. 16 B.12 C.19 D.20

32. EXPLANATION a=1 d=2 t=10 t n  = a + (n – 1) d t 10 = 1+(10 – 1) 2 = 1+ (9) 2 = 1 +18 t 10 ⇒ 19. 1,3,5,7 3-1 =2 5-3=2 7-5= 2 so, d = 2

33. 2. Find the value of n. If a = 10, d = 5, a n  = 95 A. 18 B.16 C.11 D.14

34. EXPLANATION Given, a = 10, d = 5, a n = 95 t n   = a + (n − 1) × d 95 = 10 + (n − 1) × 5 (n − 1) × 5 = 95 – 10 = 85 (n − 1) = 85/ 5 (n − 1) = 17 n = 17 + 1 ⇒ n = 18

35. 3. Find the 20th term for the given AP:3, 5, 7, 9, …… A. 11 B.41 C.21 D.26

36. EXPLANATION a = 3, d = 5 – 3 = 2, n = 20 t n = a + (n − 1) × d t 20 = 3 + (20 − 1) × 2 t 20 = 3 + 38 ⇒ t 20 = 41

37. 4. Find 16 th term in the series 7,13,19,25… A. 56 B.76 C.97 D.86

38. EXPLANATION a=7 d=6 t=16 t n  = a + (n – 1) d t 16 = 7+(16 – 1) 6 = 7+ (15) 6 = 7 +90 t 16 = 97.

39. 5. Find the no. of terms in the series 8,12,16….72 A. 14 B.17 C.27 D.36

40. EXPLANATION a=8 d=4 l=72 i.e.72-8= 64 t n  = a + (n – 1) d 72= 8+(n-1) 4 64=(n-1) 4 (n-1) = 64/4 =16 n-1 =16 n= 16+1= 17 n=17.

41. 6. Find 4 + 7 +10 + 13 + 16 +…up to 20 terms A. 650 B.630 C.324 D.506

42. EXPLANATION n = 20 a= 4 d=3 S n = n/2[2a + (n − 1) × d] = 20/2 [2(4) + (20-1) 3] = 10 [8+(19)3] = 650.

43. 7.Find 6 + 9 +12 + …. up to 30 terms A. 216 B.130 C.162 D.300

44. EXPLANATION n= 30 a=6 d=3 t n  = a + (n – 1) d 30=6+(n-1)3 24/3 = (n-1) n = 9. S n = n/2(a + l) S 9 = n/2(a + l) = 9/2(6+ 30) = 9/2 (36) S 9 = 162.

45. 8. Find the sum of first 30 multiples of 4. A. 1216 B. 1860 C.1162 D.1300

46. EXPLANATION  a = 4, n = 30, d = 4 S n = n/2 [2a + (n − 1) × d] = 30/2[2 (4) + (30 − 1) × 4] = 15[8 + 116] = 1860

47. Example 1 : 1+3.5+6+8.5+....+101

48. Find out the sum of this sequence….. 10,15,20,25,30,…………..,100

49. n=19 S n =1045

50. GEOMETRIC PROGRESSION What is a Geometric Progression? Geometric Progression (GP) is a type of sequence where each succeeding term is produced by multiplying each preceding term by a fixed number, which is called a common ratio. Example The example of GP is: 3, 6, 12, 24, 48, 96,…

51. Find the term t n = ar n-1 Formula for sum of first n terms in an GP if r >1   Sn  = a (r n  -1)/ r-1,                     if r < 1   Sn  = a (1 - r n  )/ 1-r,

52. CONTS… a = First term r = common ratio t n = nth term

53. Question 1: Find the 10 th term in the series 2,4,8,16… A. 1024 B.1030 C.1602 D.1300

54. EXPLANATION a= 2 r= 4/2= 2 n= 10 t n = ar n-1 t 10 = 2 x 2 10-1 = 1024.

55. 2. If 2,4,8,…., is the GP, then find its 10th term. A. 4 B.3 C.6 D.5

56. EXPLANATION a = 2 r = 4/2 = 2 n=10 t n = ar n-1 t 10 = 2 x 2 2-1 = 2 x 2 = 4

57. Find the sum of the first 5 terms of the following series. Given that the series is finite: 3, 6, 12, …

58. if r >1   Sn  = a (r n  -1)/ r-1,                     S 5  = 3(2 5  – 1)/(5 -1) = 93/4   

59. 3. Find 4 + 12 + 36 +… up to 6 terms A. 1216 B. 1860 C.1162 D. 1456

60. EXPLANATION a= 4 r= 3 n=6 S n = a[(r n -1)/(r-1)] r > 1 = 4(3 6 -1)/ (3-1) = 2(3 6 -1) = 1456.

61. 4. Find 1 + 1/2 + 1/4 + ….up to 5 terms

62. EXPLANATION a = 1 r = ½ n= 5 S n = a[(1- r n )/(1- r)] r < 1 S 5 = a[(1- r n )/(1- r)] = 1[1- (1/2) 5 ]/ 1-1/2 = (1- 1/32)/ 1/2 = 31/32 / ½ = 31/16 = 1 15/16.

63. 5. Find 1 + 1/2 + 1/4 + 1/8 +….∞ A. 8 B.2 C.3 D.5

64. EXPLANATION a =1 r= ½ S∞=a/1-r (if -1<r<1) S∞= 1/(1-1/2) = 2.

65. THANK YOU…