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. _I012 Harmonic Progression
A sequence is called a harmonic progression (H.P.) if the reciprocals of its terms form an A.P.
First term
General Term
It is usual to solve problems in H.P. by taking reciprocal of each term and then solving for the resulting A.P.

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

6. Harmonic Mean – a Definition
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.
Explain that formula of H.M.s is not required to be memorised even though it is in the funda book. This is because it is easier to take the corresponding AP and find the H.M.s

7. _I014 Illustrative Problem 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.
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.
Proof :
But A, G, H are in G.P.,

10. Sum of First n Natural Numbers
1, 2, 3, 4, 5,
We see that this is an A.P, with
a = 1 and d = 1.
 sum of first n Natural numbers
Explain the sigma notation to the students

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

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

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

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

15. _I017 Method of Difference 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.
You can omitt these two slides without loss of continuity

16. _I017 Method of Difference
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.
Similarly, we can define sequences of higher order of differences.

17. _I017 Method of Difference
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. _I017 Illustrative Problem 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 =  tn = 3-3n-2n2+n3

19. _I017 Method of Difference
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. _I017 Illustrative Problem Q. Find the nth term of the sequence
10, 23, 60, 169,
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. _I017 Illustrative Problem
Q. The 99th term of the series is equal to :
(a) 9998 (b) 10002
(c) 9801 (d) None of these

22. _I017 Illustrative Problem
Q. The 99th term of the series is equal to :
(a) 9998 (b) 10002
(c) 9801 (d) None of these
A. To find the 99th term of the series
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. _I017 Illustrative Problem t1 = 2 = A1+A2+A3 t2 = 7 = A1+2A2+4A3
Q. The 99th term of the series 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 =
= 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. _I014 Class Exercise Q1. A. Let the two numbers be a and b
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. _I014 Class Exercise Q2. A. Let the two quantities be a and ar2.
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. _I014 Class Exercise Q2.  Ans : (d)
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
Ask students why we have two values, and which one we will choose and why.
 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. _I015 Class Exercise Q3. A. Let the two numbers be a and b.
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. _I015 Class Exercise Q3.  Ans : (b)
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. _I015 Class Exercise Q4. 1, H1, H2, 27 are in H.P. Q.E.D.
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. _I015 Class Exercise Q5. A. Let the numbers be a and b. Given that,
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
by observation we can see that values are 3 and 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. _I016 Class Exercise Q6. A. Required sum
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. _I016 Class Exercise Q6.  Ans : (b)
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. _I016 Class Exercise Q7. A. Required sum
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

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

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

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

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

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