1 / 63

Sequences and Series

Session MPTCP04. Sequences and Series. Session Objectives. Finite and infinite sequences Arithmetic Progression (A.P.) - definition, n th term Sum of n terms of an A.P. Arithmetic Mean (A.M.) and insertion of n A.M.s between two given numbers.

cortez
Download Presentation

Sequences and Series

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Session MPTCP04 Sequences and Series

  2. Session Objectives • Finite and infinite sequences • Arithmetic Progression (A.P.) - definition, nth term • Sum of n terms of an A.P. • Arithmetic Mean (A.M.) and insertion of n A.M.s between two given numbers. • Geometric Progression (G.P.) - definition, nth term • Sum of n terms of a G.P.

  3. Sequence – a Definition _I001 A sequence is a function whose domain is the set N of natural numbers. a1, a2, a3, . . ., an, . . .

  4. Finite and Infinite Sequences _I001 Finite sequence : a  a1, a2, a3, . . ., an Infinite sequence : a  a1, a2, a3, . . ., an, . . . 

  5. Series – a Definition _I001 If a1, a2, a3, . . ., an, . . . is a sequence, the expression a1+a2+a3+ . . . +an+ . . . is called a series.

  6. Arithmetic Progression _I002 A sequence is called an arithmetic progression (A.P.) if the difference between any term and the previous term is constant. The constant difference, generally denoted by d is called the common difference. a1 = a a2 = a+d a3 = a+d+d = a+2d a4 = a+d+d+d = a+3d First term General Term an = a+d+d+d+... = a+(n-1)d

  7. Is a Given Sequence an A.P.? _I002 Algorithm to determine whether a given sequence is an A.P. : Step I Obtain general term an Step II Determine an+1 by replacing n by n+1 in the general term Step III Find an+1-an. If this is independent of n, the given sequence is an A.P.

  8. Problem Solving Tip _I002 Choose Well!!!! # Terms Common diff. 3 a-d, a, a+d d 4 a-3d, a-d, a+d, a+3d 2d 5 a-2d, a-d, a, a+d, a+2d d 6 a-5d, a-3d, a-d, a+d, a+3d, a+5d 2d

  9. Illustrative problem _I002 Q. If sum of three numbers in A.P. is 45, and the second number is thrice the first number, find the three numbers. A. Let the numbers be a-d, a, a+d Given that (a-d)+a+(a+d) = 45  3a = 45  a = 15 Also, a = 3(a-d)  3d = 30  d = 10  the three numbers are 5, 15, 25

  10. Important Properties of A.P.s _I002 If A  a, a+d, . . ., a+(n-1)d adding constant k to each term, we get, A’  a+k, a+d+k, . . ., a+(n-1)d+k A’ is also an A.P. with the same common difference.

  11. Important Properties of A.P.s _I002 If A  a, a+d, . . ., a+(n-1)d multiplying each term by non-zero constant k, A’  ak, ak+dk, . . ., ak+(n-1)dk A’ is also an A.P. with common difference dk

  12. Important Properties of A.P.s _I002 ak+an-(k-1) = a1+an  k = 2, 3, 4, . . . (n-1) Example : Consider A  2, 4, 6, 8, 10, 12, 14, 16, 18, 20

  13. Important Properties of A.P.s _I002 ak+an-(k-1) = a1+an  k = 2, 3, 4, . . . (n-1) Example : Consider A  2, 4, 6, 8, 10, 12, 14, 16, 18, 20 a3+a8 = 22

  14. Important Properties of A.P.S _I002 ak+an-(k-1) = a1+an  k = 2, 3, 4, . . . (n-1) Example : Consider A  2, 4, 6, 8, 10, 12, 14, 16, 18, 20 a3+a8 = 22 = a5+a6 = 22

  15. Important Properties of A.P.s _I002 ak+an-(k-1) = a1+an  k = 2, 3, 4, . . . (n-1) Example : Consider A  2, 4, 6, 8, 10, 12, 14, 16, 18, 20 a3+a8 = 22 = a5+a6 = 22 = a1+a10 = 22

  16. Important Properties of A.P.s _I002 a, b, c are in A.P.  2b = a+c

  17. Important Properties of A.P.s _I002 A sequence is an A.P.  an = An+B, A, B are constants. A is the common difference. Proof : an = a+(n-1)d or, an = dn+(a-d) or, an = An+B, where A is the common difference

  18. Important Properties of A.P.s _I002 If A  a, a+d, . . ., a+(n-1)d take every third term, A’  a, a+3d, a+6d, . . . . . . . . . . A’ is also an A.P.

  19. Sum of n Terms of an A.P. _I003 Sn = a1 +(a1+d)+ . . .+{a1+(n-2)d}+{a1+(n-1)d} Also, Sn = {a1+(n-1)d}+{a1+(n-2)d}+{a1+d}+. . .+a1 Adding, 2Sn = n{2a1+(n-1)d}

  20. Sum of n Terms of an A.P. _I003 This can also be written as :

  21. We know that, Rearranging, Property of Sum of n Terms of an A.P. _I003 A sequence is an A.P.  Sn = An2+Bn, where A, B are constants. 2A is the common difference. Or, Sn = An2+Bn.

  22. Arithmetic Mean _I004 A is the A.M. of two numbers a and b  a, A, and b are in A.P.  A-a = b-A  2A = a+b

  23. -4 -2 0 2 4 6 -6 A1 A2 A3 A4 A5 b a Arithmetic Mean – a Definition _I004 If n terms A1, A2, A3, . . . An are inserted between two numbers a and b such that a, A1, A2, A3, . . . , An, b form an A.P., then A1, A2, A3, . . . , An are called arithmetic means (A.M.s) of a and b.

  24. -4 -2 0 2 4 6 -6 A1 A2 A3 A4 A5 b a Arithmetic Mean – Common Difference _I004 Let n A.M.s be inserted between two numbers a and b The A.P. thus formed will have (n+2) terms. Let the common difference be d Now b = a+(n+2-1)d = a+(n+1)d

  25. Property of A.M.s _I004 Let n A.M.s A1, A2, A3, . . ., An be inserted between a and b. Then,

  26. A. Let the required A.M.s be A1, A2 and A3. Common difference d = Illustrative Problem _I004 Q. Insert 3 A.M.s between -4 and 3

  27. Geometric Progression _I005 Consider a family where every female of each generation has exactly 2 daughters. It is then possible to determine the number of females in each generation if the generation number is known. 1st Generation 1 female 2nd Generation 2 females 3rd Generation 4 females Such a progression is a Geometric Progression (G.P.)

  28. Geometric Progression _I005 A sequence is called a geometric progression (G.P.) if the ratio between any term and the previous term is constant. The constant ratio, generally denoted by r is called the common ratio. a1 = a a2 = ar a3 = ar2 a4 = ar3 First term General Term an = ar(n-1)

  29. Problem Solving Tip _I005 Choose Well!!!! # Terms Common ratio 3 a/r, a, ar r 4 a/r3, a/r, ar, ar3 r2 5 a/r2, a/r, a, ar, ar2 r 6 a/r5, a/r3, a/r, ar, ar3, ar5 r2

  30. Important Properties of G.P.s _I005 If G  a, ar, ar2, . . ., arn-1 multiplying each term by non-zero constant k, G’  ka, kar, kar2, . . ., karn-1 G’ is also a G.P. with the same common ratio.

  31. Important Properties of G.P.s _I005 If G  a, ar, ar2, . . ., arn-1 taking reciprocal of each term, G’ is also a G.P. with a reciprocal common ratio.

  32. Important Properties of G.P.s _I005 If G  a, ar, ar2, . . ., arn-1 raising each term to power k, G’  ak, akrk, akr2k, . . ., akr(n-1)k G’ is also a G.P. with common ratio rk.

  33. Important Properties of G.P.s _I005 akan-(k-1) = a1an  k = 2, 3, 4, . . . (n-1) Example : Consider G  1, 2, 4, 8, 16, 32, 64, 128, 256, 512

  34. Important Properties of G.P.s _I005 akan-(k-1) = a1an  k = 2, 3, 4, . . . (n-1) Example : Consider G  1, 2, 4, 8, 16, 32, 64, 128, 256, 512 a3a8 = 512

  35. Important Properties of G.P.s _I005 akan-(k-1) = a1an  k = 2, 3, 4, . . . (n-1) Example : Consider G  1, 2, 4, 8, 16, 32, 64, 128, 256, 512 a3a8 = 512 = a5a6 = 512

  36. Important Properties of G.P.s _I005 akan-(k-1) = a1an  k = 2, 3, 4, . . . (n-1) Example : Consider G  1, 2, 4, 8, 16, 32, 64, 128, 256, 512 a3a8 = 512 = a5a6 = 512 = a1a10 = 512

  37. Important Properties of G.P.s _I005 a, b, c are in G.P.  b2 = ac

  38. Important Properties of G.P.s _I005 If G  a, ar, ar2, . . ., arn-1 take every third term, G’  a, ar3, ar6, . . . G’ is also a G.P.

  39. Important Properties of G.P.s _I005 a1, a2, a3, . . . , an is a G.P. of positive terms  loga1, loga2, loga3, . . . logan is an A.P.

  40. Sum of n Terms of a G.P. _I006 Sn = a+ar+ar2+ar3+ . . .+ar(n-1) ………(i) Multiplying by r, we get, rSn = ar+ar2+ar3+ . . .+ar(n-1)+arn ……...(ii) Subtracting (i) from (ii), (r-1)Sn = a(rn-1)

  41. Class Exercise Q1. _I002 Q. If log 2, log (2x-1) and log (2x+3) are in A.P., find x.

  42. Class Exercise Q1. _I002 Q. If log 2, log (2x-1) and log (2x+3) are in A.P., find x. A. Given that log(2x-1)-log2 = log(2x+3)-log(2x-1)

  43. Class Exercise Q2. _I002 Q. Show that there is no infinite A.P. which consists only of distinct primes.

  44. Class Exercise Q2. _I002 Q. Show that there is no infinite A.P. which consists only of distinct primes. A. Let, if possible, there be an A.P. consisting only of distinct primes : a1, a2, a3, . . ., an, . . . an = a1+(n-1)d Thus, (a1+1)th term is a multiple of a1. Thus, no such A.P. is possible. Q.E.D.

  45. Q. , where Sn denotes the sum of the first n terms of an A.P., then common difference is : (a) P+Q (b) 2P+3Q (c) 2Q (d) Q (J.E.E. West Bengal 1994) Class Exercise Q3. _I003

  46. Q. , where Sn denotes the sum of the first n terms of an A.P., then common difference is : (a) P+Q (b) 2P+3Q (c) 2Q (d) Q (J.E.E. West Bengal 1994) Class Exercise Q3. _I003 A. an = Sn - Sn-1  Ans : (d).

  47. Class Exercise Q4. _I003 Q. If 12th term of an A.P. is -13 and the sum of the first four terms is 24, what is the sum of first 10 terms?

  48. Class Exercise Q4. _I003 Q. If 12th term of an A.P. is -13 and the sum of the first four terms is 24, what is the sum of first 10 terms? A. Given that, a12 = a1+11d = -13 . . . (i) S4 = 2(2a1+3d) = 24 . . . (ii) Solving (i) and (ii) simultaneously, we get, a1 = 9, d = -2  S10 = 5(2a1+9d) = 5(18-18) = 0

  49. Q. Find the value of n so that be an A.M. between a and b (a, b are positive). Class Exercise Q5. _I004

  50. Q. Find the value of n so that be an A.M. between a and b (a, b are positive). Class Exercise Q5. _I004 A. Given that, Dividing throughout by bn+1, we get,

More Related