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Session MPTCP06

Session MPTCP06. Sequences and Series. Session Objectives. Revisit H.P. Sum of n terms of an H.P. Harmonic Mean (H.M.) and insertion of n H.M.s between two given numbers Relation between A.M., G.M. and H.M. n, n 2 , n 3 Method of difference in special sequences to find general term.

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Session MPTCP06

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  1. Session MPTCP06 Sequences and Series

  2. Session Objectives • Revisit H.P. • Sum of n terms of an H.P. • Harmonic Mean (H.M.) and insertion of n H.M.s between two given numbers • Relation between A.M., G.M. and H.M. • n, n2, n3 • Method of difference in special sequences to find general term

  3. Harmonic Progression _I012 A sequence is called a harmonic progression (H.P.) if the reciprocals of its terms form an A.P. First term General Term

  4. Sum of n Terms of an H.P. _I013 There is no general formula for sum of n terms of an H.P.

  5. H is the H.M. of a and b  a, H, and b are in H.P.  are in A.P. Single Harmonic Mean _I014

  6. Harmonic Mean – a Definition _I014 If n terms H1, H2, H3, . . . Hn are inserted between two numbers a and b such that a, H1, H2, H3, . . . , Hn, b form an H.P., then H1, H2, H3, . . . , Hn are called harmonic means (H.M.s) of a and b.

  7. Illustrative Problem _I014 Q. Insert 3 H.M.s between 1 and 1/9 A. Let the required H.M.s be H1, H2 and H3.

  8. Relation Between A.M., G.M. And H.M. _I015 Let A, G and H be the arithmetic, geometric and harmonic means of two positive numbers a and b. G2=AH Proof :

  9. Relation Between A.M., G.M. And H.M. _I015 Proof : But A, G, H are in G.P.,

  10. Sum of First n Natural Numbers _I016 First n Natural numbers : 1, 2, 3, 4, 5, 6 . . . We see that this is an A.P, with a = 1 and d = 1.  sum of first n Natural numbers

  11. Sum of Squares of First n Natural Numbers _I016 Squares of first n Natural numbers : 12, 22, 32, 42, 52, 62 . . . Consider the identity : Putting x = 1, 2, 3, . . . n, we get

  12. Sum of Squares of First n Natural Numbers _I016 Adding columnwise, we get,

  13. Sum of Cubes of First n Natural Numbers (H.W) _I016 Cubes of first n Natural numbers: 13, 23, 33, 43, 53, 63 . . . Consider the identity : Putting x = 1, 2, 3, . . . n, we get

  14. Sum of Cubes of First n Natural Numbers (H.W.) _I016 Adding columnwise, we get,

  15. Method of Difference _I017 Consider the sequence : U  u1, u2, u3 . . If we define 1ur = ur+1-ur, then the sequence 1U 1u1, 1u2, 1u3, . . . 1un-1 is called the sequence of the first order of differences.

  16. Method of Difference _I017 Similarly, if we define 2ur= 1ur+1-1ur, then the sequence, 2U 2u1, 2u2, 2u3, . . . 2un-2 is called the sequence of the second order of differences.

  17. Method of Difference _I017 If the sequence of the first order of differences is an A.P., the general term of the sequence is of the form tn = A1+A2n+A3n2 If the sequence of the second order of differences is an A.P., the general term of the sequence is of the form tn = A1+A2n+A3n2+A4n3

  18. Illustrative Problem _I017 Q. Find the nth term of the sequence -1, -3, 3, 23, 63, 129, . . . A. We see that the sequence of the second order of differences is an A.P. Therefore the general term is : tn = A1+A2n+A3n2+A4n3 t1 = -1 = A1+A2+A3+A4; t2 = -3 = A1+2A2+4A3+8A4; t3 = 3 = A1+3A2+9A3+27A4; t4 = 23 = A1+4A2+16A3+64A4  A1 = 3, A2 = -3, A3 = -2, A4 = 1  tn = 3-3n-2n2+n3

  19. Method of Difference _I017 If the sequence of the second order of differences is a G.P., in which the common ratio is r, the general term of the sequence is of the form tn = A1+A2n+A3rn-1

  20. Illustrative Problem _I017 Q. Find the nth term of the sequence 10, 23, 60, 169, 494 . . . A. We see that the sequence of the second order of differences is a G.P. with common ratio 3. Therefore the general term is : tn = A1+A2n+A33n-1 t1 = 10 = A1+A2+A3; t2 = 23 = A1+2A2+3A3; t3 = 60 = A1+3A2+9A3;  A1 = 3, A2 = 1, A3 = 6  tn = 3+n+6(3n-1)= 3+n+2(3n)

  21. Illustrative Problem _I017 Q. The 99th term of the series 2+7+14+23+34. . . is equal to : (a) 9998 (b) 10002 (c) 9801 (d) None of these

  22. Illustrative Problem _I017 Q. The 99th term of the series 2+7+14+23+34. . . is equal to : (a) 9998 (b) 10002 (c) 9801 (d) None of these A. To find the 99th term of the series 2+7+14+23+34+ . . . We see that the sequence of the first order of difference is an A.P.  the nth term is given by tn = A1+A2n+A3n2

  23. Illustrative Problem _I017 Q. The 99th term of the series 2+7+14+23+34. . . is equal to : (a) 9998 (b) 10002 (c) 9801 (d) None of these tn = A1+A2n+A3n2 t1 = 2 = A1+A2+A3 t2 = 7 = A1+2A2+4A3 t3 = 14 = A1+3A2+9A3  A1 = -1; A2 = 2; A3 = 1.  tn = n2+2n-1  t99 = 992+2(99)-1 = 9801+198-1 = 9998  Ans : (a)

  24. Home work Find

  25. Class Exercise Q1. _I014 Q. The H.M. of 2 numbers is 4. Their A.M. A and G.M. G satisfy the relation 2A+G2 = 27. Find the numbers.

  26. Class Exercise Q1. _I014 Q. The H.M. of 2 numbers is 4. Their A.M. A and G.M. G satisfy the relation 2A+G2 = 27. Find the numbers. A. Let the two numbers be a and b Given that 2A+G2 = 27  a+b+ab = 27  3(a+b) = 27  a+b = 9, ab = 18 Now, a2+2ab+b2 = 81,  a2-2ab+b2 = 9  a-b =  3  a = 6, b = 3 or a = 3, b = 6

  27. Class Exercise Q2. _I014 Q. If the harmonic mean between two quantities is to their geometric mean as 12 to 13, the quantities are in the ratio of (a) 1:2 (b) 2:3 (c) 3:5 (d) 4:9

  28. Class Exercise Q2. _I014 Q. If the harmonic mean between two quantities is to their geometric mean as 12 to 13, the quantities are in the ratio of (a) 1:2 (b) 2:3 (c) 3:5 (d) 4:9 A. Let the two quantities be a and ar2. Given that

  29. Class Exercise Q2. _I014 Q. If the harmonic mean between two quantities is to their geometric mean as 12 to 13, the quantities are in the ratio of (a) 1:2 (b) 2:3 (c) 3:5 (d) 4:9  Ans : (d)

  30. Class Exercise Q3. _I015 Q. If between any two quantities there be inserted two A.M.s A1, A2; two G.M.s G1, G2; and two H.M.s H1, H2, then A1+A2:H1+H2 is equal to : (a) H1H2:G1G2 (b) G1G2:H1H2 (c) H1H2:A1A2 (d) G1G2:A1A2

  31. Class Exercise Q3. _I015 Q. If between any two quantities there be inserted two A.M.s A1, A2; two G.M.s G1, G2; and two H.M.s H1, H2, then A1+A2:H1+H2 is equal to : (a) H1H2:G1G2 (b) G1G2:H1H2 (c) H1H2:A1A2 (d) G1G2:A1A2 A. Let the two numbers be a and b.  A1+A2 = a+b  G1G2 = ab and

  32. Class Exercise Q3. _I015 Q. If between any two quantities there be inserted two A.M.s A1, A2; two G.M.s G1, G2; and two H.M.s H1, H2, then A1+A2:H1+H2 is equal to : (a) H1H2:G1G2 (b) G1G2:H1H2 (c) H1H2:A1A2 (d) G1G2:A1A2  Ans : (b)

  33. Class Exercise Q4. _I015 Q. Insert 2 A.M.s, 2 G.M.s and 2 H.M.s between 1 and 27. Hence show that corresponding A.M. > G.M. > H.M.

  34. Class Exercise Q4. _I015 Q. Insert 2 A.M.s, 2 G.M.s and 2 H.M.s between 1 and 27. Hence show that corresponding A.M. > G.M. > H.M. A. Let A.M.s, G.M.s and H.M.s of 1 and 27 be A1, A2, G1, G2 and H1, H2 respectively. 1, A1, A2, 27 are in A.P. 1, G1, G2, 27 are in G.P.  G1 = 3, G2 = 9

  35. Class Exercise Q4. _I015 Q. Insert 2 A.M.s, 2 G.M.s and 2 H.M.s between 1 and 27. Hence show that corresponding A.M. > G.M. > H.M.  G1 = 3, G2 = 9 1, H1, H2, 27 are in H.P. Q.E.D.

  36. Class Exercise Q5. _I015 Q. The G.M. of two numbers is 9 and their A.M. is 15. Find the numbers.

  37. Class Exercise Q5. _I015 Q. The G.M. of two numbers is 9 and their A.M. is 15. Find the numbers. A. Let the numbers be a and b. Given that,  a+b = 30, ab = 81  a2+2ab+b2 = 900  a2-2ab+b2 = 576  a-b = 24  a = 27, b = 3 or a = 3, b = 27

  38. Class Exercise Q6. _I016 Q. Sum of n terms of a series whose nth term is 3(4n+2n2)-4n3 is equal to (a) 4n+1-4+n(n+1)(1+n+n2) (b) 4n+1-4+n(n-1)(1+n-n2) (c) 4n+1-4+n(n+1)(1+n-n2) (d) None of these

  39. Class Exercise Q6. _I016 Q. Sum of n terms of a series whose nth term is 3(4n+2n2)-4n3 is equal to (a) 4n+1-4+n(n+1)(1+n+n2) (b) 4n+1-4+n(n-1)(1+n-n2) (c) 4n+1-4+n(n+1)(1+n-n2) (d) None of these A. Required sum

  40. Class Exercise Q6. _I016 Q. Sum of n terms of a series whose nth term is 3(4n+2n2)-4n3 is equal to (a) 4n+1-4+n(n+1)(1+n+n2) (b) 4n+1-4+n(n-1)(1+n-n2) (c) 4n+1-4+n(n+1)(1+n-n2) (d) None of these  Ans : (b)

  41. Class Exercise Q7. _I016 Q. Sum of n terms of a series whose nth term is 2n-1+8n3-6n2 is equal to (a) 2n-1+n(n+1)(2n2-1) (b) 2n-1+n(n+1)(2n2+1) (c) 2n+1+n(n+1)(2n2-1) (d) None of these

  42. Class Exercise Q7. _I016 Q. Sum of n terms of a series whose nth term is 2n-1+8n3-6n2 is equal to (a) 2n-1+n(n+1)(2n2-1) (b) 2n-1+n(n+1)(2n2+1) (c) 2n+1+n(n+1)(2n2-1) (d) None of these A. Required sum

  43. Class Exercise Q7. _I016 Q. Sum of n terms of a series whose nth term is 2n-1+8n3-6n2 is equal to (a) 2n-1+n(n+1)(2n2-1) (b) 2n-1+n(n+1)(2n2+1) (c) 2n+1+n(n+1)(2n2-1) (d) None of these

  44. Class Exercise Q8. _I016 Q. Sum the series 43+53+63+ . . . +203

  45. Class Exercise Q8. _I016 Q. Sum the series 43+53+63+ . . . +203 A. Sn = 43+53+63+ . . . +203

  46. Class Exercise Q9(i). _I016 Q. Sum to n terms the series : 1.7+2.8+3.9+ . . .

  47. Class Exercise Q9(i). _I016 Q. Sum to n terms the series : 1.7+2.8+3.9+ . . . A. We see that an = n(n+6)

  48. Class Exercise Q9(ii). _I016 Q. Sum to n terms the series : 1.3+3.5+5.7+ . . .

  49. Class Exercise Q9(ii). _I016 Q. Sum to n terms the series : 1.3+3.5+5.7+ . . . A. We see that an = (2n-1)(2n+1) = 4n2-1

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